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CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Vector Algebra
Quiz 11
Let $$ABC$$ be a triangle, the position vector of whose vertices are $$7 \hat j + 10 \hat k, - \hat i + 6 \hat j + 6 \hat k$$ and $$- 4 \hat i + 9 \hat j + 6 \hat k$$. Then $$\Delta ABC$$ is
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isosceles
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equilateral
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right-angled
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none of these
Explanation
We have, $$\overrightarrow{AB} = - \hat i - \hat j - 4 \hat k, \overrightarrow{BC} = - 3 \hat i + 3 \hat j$$ and $$\overrightarrow{CA} = 4 \hat i - 2 \hat j + 4 \hat k$$. Therefore,
$$\mid \overrightarrow{AB}\mid = \mid \overrightarrow{BC}\mid = 3 \sqrt 2$$ and $$\mid \overrightarrow{CA}\mid = 6$$
Clearly, $$\mid \overrightarrow{AB}\mid^2 = \mid \overrightarrow{BC}\mid^2 = \mid \overrightarrow{AC}\mid^2$$
Hence, the triangle s right-angled isosceles triangle.
Vector $$ \vec{x} $$ is
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$$\dfrac{1}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times(\vec{a} \times \vec{b})]$$
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$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$$
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$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$$
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none of these
Explanation
Given
$$ \vec{x} \times \vec{y}=\vec{a} $$ $$\qquad (i)$$
$$ \vec{y} \times \vec{z}=\vec{b} $$ $$\qquad (ii)$$
$$ \vec{x} \cdot \vec{b}=\gamma $$ $$\qquad (iii)$$
$$ \vec{x} \cdot \vec{y}=1 $$ $$\qquad (iv)$$
$$ \vec{y} \cdot \vec{z}=1 $$ $$\qquad (v)$$
From (ii), $$ \vec{x} \cdot(\vec{y} \times \vec{z})=\vec{x} \cdot \vec{b}=\gamma \Rightarrow[\vec{x} \vec{y} \vec{z}]=\gamma $$
From (i) and (ii), $$ (\vec{x} \times \vec{y}) \times(\vec{y} \times \vec{z})=\vec{a} \times \vec{b} $$
$$ \therefore[\vec{x} \vec{y} \vec{z}] \vec{y}-[\vec{y} \vec{y} \vec{z}] \vec{x}=\vec{a} \times \vec{b} \Rightarrow \vec{y}=\dfrac{\vec{a} \times \vec{b}}{\gamma} $$ $$\qquad (vi)$$
Also from (i), we get $$ (\vec{x} \times \vec{y}) \times \vec{y}=\vec{a} \times \vec{y} $$
$$ \Rightarrow(\vec{x} \cdot \vec{y}) \vec{y}-(\vec{y} \cdot \vec{y}) \vec{x}=\vec{a} \times \vec{y} \Rightarrow \vec{x}=\left(1 /|\vec{y}|^{2}\right)(\vec{y}-\vec{a} \times \vec{y})=\dfrac{\gamma^{2}}{|\vec{a} \times \vec{b}|^{2}}\left[\dfrac{\vec{a} \times \vec{b}}{\gamma}-\dfrac{\vec{a} \times(\vec{a} \times \vec{b})}{\gamma}\right] $$
$$\begin{array}{l}\Rightarrow \vec{x}=\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}-\vec{a} \times(\vec{a} \times \vec{b})] \end{array}$$
Vector $$ \vec{z} $$ is
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$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b} \times(\vec{a} \times \vec{b})] $$
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$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$$
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$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$$
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none of these
Explanation
Given
$$ \vec{x} \times \vec{y}=\vec{a} $$ $$\qquad (i)$$
$$ \vec{y} \times \vec{z}=\vec{b} $$ $$\qquad (ii)$$
$$ \vec{x} \cdot \vec{b}=\gamma $$ $$\qquad (iii)$$
$$ \vec{x} \cdot \vec{y}=1 $$ $$\qquad (iv)$$
$$ \vec{y} \cdot \vec{z}=1 $$ $$\qquad (v)$$
$$\begin{array}{l} \text { Also from (ii), }(\vec{y} \times \vec{z}) \times \vec{y}=\vec{b} \times \vec{y} \Rightarrow|\vec{y}|^{2} \vec{z}-(\vec{z} \cdot \vec{y}) \vec{y}=\vec{b} \times \vec{y} \\\Rightarrow \vec{z}=\dfrac{1}{|\vec{y}|^{2}}[\vec{y}+\vec{b} \times \vec{y}]=\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]\end{array}$$
Vector $$ \vec{y} $$ is
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$$\dfrac{\vec{a} \times \vec{b}}{\gamma}$$
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$$\vec{a}+\dfrac{\vec{a} \times \vec{b}}{\gamma}$$
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$$\vec{a}+\vec{b}+\dfrac{\vec{a} \times \vec{b}}{\gamma} $$
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none of these
Explanation
Given
$$ \vec{x} \times \vec{y}=\vec{a} $$ $$\qquad (i)$$
$$ \vec{y} \times \vec{z}=\vec{b} $$ $$\qquad (ii)$$
$$ \vec{x} \cdot \vec{b}=\gamma $$ $$\qquad (iii)$$
$$ \vec{x} \cdot \vec{y}=1 $$ $$\qquad (iv)$$
$$ \vec{y} \cdot \vec{z}=1 $$ $$\qquad (v)$$
From (ii), $$ \vec{x} \cdot(\vec{y} \times \vec{z})=\vec{x} \cdot \vec{b}=\gamma \Rightarrow[\vec{x} \vec{y} \vec{z}]=\gamma $$
From (i) and (ii), $$ (\vec{x} \times \vec{y}) \times(\vec{y} \times \vec{z})=\vec{a} \times \vec{b} $$
$$ \therefore[\vec{x} \vec{y} \vec{z}] \vec{y}-[\vec{y} \vec{y} \vec{z}] \vec{x}=\vec{a} \times \vec{b} \Rightarrow \vec{y}=\dfrac{\vec{a} \times \vec{b}}{\gamma} $$
Vectors $$\vec{A} \space and \space \vec{B}$$ satisfying the vector equation $$\vec{A} + \vec{B} = \vec{a}, \vec{A} \times \vec{B} = \vec{b} \space and \space \vec{A} . \vec{a} = 1$$ where $$\vec{a} \space and \space \vec{b}$$ are given vectors are
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$$\vec{A} = \frac{(\vec{a} \times \vec{b}) - \vec{a}}{a^2}$$
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$$\vec{B} = \frac{(\vec{b} \times \vec{a}) + \vec{a}(a^2 - 1)}{a^2}$$
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$$\vec{A} = \frac{(\vec{a} \times \vec{b}) + \vec{a}}{a^2}$$
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$$\vec{B} = \frac{(\vec{b} \times \vec{a}) - \vec{a}(a^2 - 1)}{a^2}$$
$$ (\vec{P} \times \vec{B}) \times \vec{B} $$ is equal to
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$$\vec{P}$$
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$$-\vec{P}$$
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$$2 \vec{B}$$
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$$\vec{A}$$
Explanation
$$ \vec{P} \times \vec{B}=\vec{A}-\vec{P} $$ and $$ |\vec{A}|=|\vec{B}|=1 $$ and $$ \vec{A} \cdot \vec{B}=0 $$ is given
Now $$ \vec{P} \times \vec{B}=\vec{A}-\vec{P} $$ $$ \qquad (i)$$
$$ (\vec{P} \times \vec{B}) \times \vec{B}=(\vec{A}-\vec{P}) \times \vec{B} $$ (taking cross product with $$ \vec{B} $$ on both sides)
$$\begin{array}{l}\Rightarrow(\vec{P} \cdot \vec{B}) \vec{B}-(\vec{B} \cdot \vec{B}) \vec{P}=\vec{A} \times \vec{B}-\vec{P} \times \vec{B} \\\Rightarrow(\vec{P} \cdot \vec{B}) \vec{B}-\vec{P}=\vec{A} \times \vec{B}-\vec{A}+\vec{P} \\\Rightarrow 2 \vec{P}=\vec{A}-\vec{A} \times \vec{B}-(\vec{P} \cdot \vec{B}) \vec{B} \\\Rightarrow \vec{P}=\dfrac{\vec{A}-\vec{A} \times \vec{B}-(\vec{P} \cdot \vec{B}) \vec{B}}{2}\end{array}$$ $$ \qquad (ii)$$
Taking dot product with $$ \vec{B} $$ on both sides of (i), we get
$$\begin{array}{l}\vec{P} \cdot \vec{B}=\vec{A} \cdot \vec{B}-\vec{P} \cdot \vec{B} \\\Rightarrow \vec{P} \cdot \vec{B}=0 \qquad \qquad (iii) \\\Rightarrow \vec{P}=\dfrac{\vec{A}+\vec{B} \times \vec{A}}{2}\end{array}$$
Now
$$(\vec{P} \times \vec{B}) \times \vec{B}=(\vec{P} \cdot \vec{B}) \vec{B}-(\vec{B} \cdot \vec{B}) \vec{P}=-\vec{P}$$
If side $$\vec{AB}$$ of an equilateral triangle ABC lying in the x - y plane is $$3\hat{i}$$, then side $$\vec{CB}$$ can be
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$$-\frac{3}{2}(\hat{i} - \sqrt{3}\hat{j}$$
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$$\frac{3}{2}(\hat{i} - \sqrt{3}\hat{j})$$
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$$-\frac{3}{2}(\hat{i} + \sqrt{3}\hat{j})$$
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$$\frac{3}{2}(\hat{i} - \sqrt{3} \hat{j})$$
Given that $$\vec a, \vec b, \vec p, \vec q$$ are four vectors such that $$\vec a + \vec b = \mu \vec p, \vec b \cdot \vec q = 0$$ and $$(\vec b)^2 = 1,$$ where $$\mu$$ is scalar. Then $$\mid (\vec a \cdot \vec q) \vec p - (\vec p \cdot \vec q)\vec a \mid$$ is equal to
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$$2 \mid \vec p \cdot \vec q \mid$$
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$$(1/2) \mid \vec p \cdot \vec q \mid$$
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$$\mid \vec p \times \vec q \mid$$
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$$\mid \vec p \cdot \vec q \mid$$
Explanation
$$\vec a + \vec b = \mu \vec p,$$ $$ \vec b \cdot \vec q = 0$$, $$(\vec b)^2 = 1,$$
$$\because \vec a + \vec b = \mu \vec p$$
$$\Rightarrow (\vec a + \vec b) \times \vec a = \mu \vec p \times \vec a, \vec b \times \vec a = \mu \vec p \times \vec a \Rightarrow \vec q \times (\vec b \times \vec a) = \mu \vec q \times (\vec p \times \vec q)$$
$$\Rightarrow (\vec q \cdot \vec a) \vec b - (\vec q \cdot \vec b) \vec a = \mu \vec q \times (\vec p \times \vec a) \Rightarrow (\vec q \cdot \vec a) \vec b = \mu \vec q \times (\vec p \times \vec a)$$
$$\because \vec a + \vec b = \mu \vec p$$
$$\Rightarrow \vec q \cdot (\vec a + \vec b) = \mu \vec q \cdot \vec p$$
$$\Rightarrow \vec q \cdot \vec a + \vec q \cdot \vec b = \mu \vec p \cdot \vec q$$
$$\mu = \dfrac{\vec q \cdot \vec a}{\vec p \cdot \vec q}$$
$$\Rightarrow (\vec q \cdot \vec a) \vec b = \dfrac{\vec q \cdot \vec a}{\vec p \cdot \vec q} [(\vec q \cdot \vec a) \cdot \vec p - (\vec q \cdot \vec p)\vec a]$$
$$\Rightarrow \mid (\vec q \cdot \vec a)\vec p - (\vec q \cdot \vec p)\vec a \mid = \mid (\vec p \cdot \vec q) \vec b \mid = \mid (\vec p \cdot \vec q)\mid \cdot \mid \vec b\mid$$
$$\Rightarrow \mid (\vec q \cdot \vec a)\vec p - (\vec q \cdot \vec p) \vec a \mid = \mid \vec p \cdot \vec q \mid$$
If $$\vec r$$ and $$\vec s$$ are non-zero constant vectors and the scalar $$b$$ is chosen such that $$\mid \vec r + b \vec s \mid$$ is minimum, then the value of $$\mid b \vec s \mid^2 + \mid \vec r + b \vec s \mid^2$$ is equal to
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$$2 \mid \vec r \mid^2$$
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$$\mid \vec r \mid^2/2$$
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$$3 \mid \vec r \mid^2$$
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$$\mid \vec r \mid^2$$
Explanation
For minimum value $$\mid \vec r + b \vec s \mid = 0$$
Let $$\vec r$$ and $$\vec s$$ are anti parallel so $$b \vec s = - \vec r$$
so $$\mid b \vec s\mid^2 + \mid \vec r + b \vec s \mid^2 = \mid - \vec r \mid^2 + \mid \vec r - \vec r \mid^2 = \mid \vec r \mid^2$$
If $$\overline a$$ and $$\overline b$$ are adjacent sides of a rhombus, then $$\overline a.\overline b=0$$.
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True
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False
Explanation
If $$\overline a.\overline b=0$$, then $$\overline a . \overline b=|\overline
a| . |\overline b| \cos 90^o$$
Hence, angle between $$\overline a$$ and $$\overline b$$ is $$90^o$$, which is not possible in a rhombus.
Since, angle between adjacent sides in a rhombus is not equal to $$90^o$$.
Let $$\hat{a} \space and \space \hat{b}$$ be mutually perpendicular unit vectors. Then for any arbitrary $$\vec{r}$$.
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$$\vec{r} = (\vec{r} . \hat{a})\hat{a} + (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$$
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$$\vec{r} = (\vec{r} . \hat{a}) - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b}))) (\hat{a} \times \hat{b})$$
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$$\vec{r} = (\vec{r} . \hat{a})\hat{a} - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$$
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none of these
The vector with initial point $$P(2,-3,5)$$ and terminal point $$Q(3,-4,7)$$ is
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$$\hat i-\hat j+2\hat k$$
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$$5\hat i-7\hat j+12\hat k$$
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$$\hat i+\hat j-2\hat k$$
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$$None\ of\ these$$
Explanation
$$(A)$$ is the correct answer.
Required vector $$=|\vec{AB}|=(3-2) \hat i+(-4+3)\hat j+(7-5)\hat k$$
$$=\hat i-\hat j+2\hat k$$
If $$ \vec{X} \cdot \vec{A}=0, \vec{X} \cdot \vec{B}=0 $$ and $$ \vec{X} \cdot \vec{C}=0 $$ for some non-zero vector $$ \vec{X}, $$ then $$ [\vec{A} \vec{B} \vec{C}]=0 $$
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True
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False
The position vector of the point which divides the join of points with position vectors $$\vec a +\vec b$$ and $$2\vec a-\vec b$$ in the ratio $$1:2$$ is
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$$\dfrac {3\vec a+2\vec b}{3}$$
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$$\vec a$$
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$$\dfrac {5\vec a-\vec b}{3}$$
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$$\dfrac {4\vec a+\vec b}{3}$$
Explanation
$$(D)$$ is the correct answer. Applying section formula, the position vector of the required point is $$\dfrac {2(\vec a+\vec b)+1(2\vec a-\vec b)}{2+1}=\dfrac {4\vec a+\vec b}{3}$$
Since, the position vector of a point $$R$$ divides the line segment joining the points $$P$$ and $$Q$$, whose position vectors are $$\vec p$$ and $$\vec q$$ in the ration $$m:n$$ internally, is given by $$\dfrac{m\vec q +n\vec p}{m+n}$$
The projection of vector $$\vec a=2\hat i-\hat j+\hat k$$ along $$\vec b=\hat i+2\hat j+2\hat k$$ is
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$$\dfrac {2}{3}$$
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$$\dfrac {1}{3}$$
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$$2$$
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$$\sqrt 6$$
Explanation
$$(A)$$ is the correct answer. Projection of a vector $$\vec a$$ on $$\vec b$$ is
$$\dfrac {\vec a \vec b}{|b|}=\dfrac {(2\hat i-\hat j+\hat k)(\hat i+2\hat j+2\hat k)}{\sqrt {1+4+4}}=\dfrac 23$$
Position vector of a point $$P$$ is a vector whose initial point is origin.
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True
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False
Explanation
Since, $$\overline P=\overline{OP}$$ displacement of vector $$\overline P$$ from origin.
Let $$ \vec{u}, \vec{v} $$ and $$ \vec{w} $$ be vectors such that $$ \vec{u}+\vec{v}+\vec{w}=0 . $$ If $$ |\vec{u}|=3,|\vec{v}|=4 $$ and $$ |\vec{w}|=5, $$ then $$ \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u} $$ is
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$$47$$
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$$-25$$
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$$0$$
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$$25$$
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$$50$$
Line $$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$$ will not meet the plane $$\overrightarrow{r} \cdot \overrightarrow{n} = q$$, if
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$$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$$
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$$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$$
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$$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$$
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$$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$$
Explanation
Given line is $$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$$
Substitute it in plane equation $$\overrightarrow{r} \cdot \overrightarrow{n} = q$$,
$$( \overrightarrow{a} + \lambda \overrightarrow{b}).\overrightarrow{n}=q$$
$$( \overrightarrow{a}. \overrightarrow{n}+ \lambda \overrightarrow{b}.\overrightarrow{n})=q$$
We must have $$\overrightarrow{b} \cdot \overrightarrow{n} = 0$$
and $$\overrightarrow{a} \cdot \overrightarrow{n} \neq q$$ (as point $$\overrightarrow{a} $$ on the line should not lie on the plane.
If $$\vec \alpha | =4$$ and $$ -3 \le \lambda \le 2$$, then the range of $$ | \lambda \vec \alpha |$$ is
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$$[0, 8]$$
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$$[-12, 8]$$
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$$[0, 12]$$
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$$[8, 12]$$
Explanation
We have, $$| \vec \alpha | =4$$ and $$-3 \le \lambda \le 2$$
$$\therefore | \lambda \vec a| =|-3| 4=12$$, at $$\lambda =-3$$
$$| \lambda \vec a| =|0|4=0$$, at $$\lambda =0$$
And $$| \lambda \vec \alpha | |2| 4=8$$, at $$\lambda =2$$
So, the range of $$| \lambda \vec \alpha |$$ is $$[0, 12]$$.
Alternate Method
Since, $$-3 \le \lambda \le 2$$
$$0 \le || \lambda | \le 3$$
$$\Rightarrow 0 \le 4 | \lambda | \le 12$$
$$| \lambda \vec a| \in [ 0, 12]$$
If $$\vec a, \vec b, \vec c$$ are unit vector such that $$\vec a +\vec b +\vec c=\vec 0$$, then the value of $$\vec a \vec b+\vec b. \vec c+\vec c. \vec a$$ is
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$$1$$
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$$3$$
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$$-\dfrac 32$$
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$$None\ of\ these$$
Explanation
We have, $$\vec a+ \vec b+ \vec c= \vec 0$$, and $$\vec a^2=1, \vec b^2 =1, \vec c^2 =1 | \vec a|=1, | \vec b|=1, | \vec c| =1$$
$$\because ( \vec a+ \vec b +\vec c)( \vec a+ \vec b +\vec c)=0$$
$$\Rightarrow \vec a^2 + \vec a. \vec b + \vec a. \vec c+ \vec +\vec b . \vec a + \vec b^2 + \vec b. \vec c+ \vec c. \vec a+ \vec c. \vec b +\vec c^2=0$$
$$\Rightarrow \vec a^2 +\vec b^2.+\vec c^2 +2( \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a)=0$$
$$[ \because \vec a. \vec b =\vec b. \vec a, \vec b. \vec c= \vec c. \vec b$$ and $$\vec c. \vec a= \vec a. \vec c]$$
$$\Rightarrow 1+1+1+2( \vec a. \vec b + \vec b. \vec c + \vec c. \vec a)=0$$
$$\Rightarrow \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a=-\dfrac 32$$
The vector having initial and terminal points as $$(2, 5, 0)$$ and $$(-3, 7, 4)$$, respectively is
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$$-\hat i +12 \hat j +4\hat k$$
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$$5\hat i+2\hat j-4\hat k$$
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$$-5\hat i +2\hat j+4\hat k$$
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$$\hat i+\hat j+ \hat k$$
Explanation
Required vector $$=|\vec{AB}|=(-3-2) \hat i+(7-5)\hat j+(4-0)\hat k$$
$$=-5\hat i+2\hat j=4\hat k$$
If $$\vec a, \vec b, \vec c$$ are three vectors such that $$ \vec a +\vec b+ \vec c=\vec 0$$ and $$ | \vec a| =2, | \vec b|=3, | \vec c| =5$$, then value of $$\vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a$$ is
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$$0$$
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$$1$$
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$$-19$$
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$$38$$
Explanation
Here, $$ \vec a+ \vec b+ \vec c=\vec 0$$ and $$\vec a^2 =4, \vec b^2 =9, \vec c^2=25$$
$$\therefore ( \vec a+ \vec b+ \vec c). ( \vec a+ \vec b+\vec c)=\vec 0$$
$$\Rightarrow \vec a^2 +\vec a. \vec b+ \vec a. \vec c+\vec b. \vec a+\vec b^2 + \vec b. \vec c+ \vec c. \vec a+ \vec c. \vec b+ \vec c^2=\vec 0$$
$$\Rightarrow \vec a^2+ \vec b^2+\vec c^2+2( \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a)=0$$ $$[ \because \vec a. \vec b=\vec b. \vec a]$$
$$\Rightarrow 4+9+25+2( \vec a. \vec b+ \vec b. \vec c +\vec c. \vec a)=0$$
$$\Rightarrow \vec a. \vec b+ \vec b. \vec c+ \vec c. \vec a=\dfrac{-38}{2}=-19$$
The position vector of the point which divides the join of points $$2 \vec a -3\vec b$$ and $$\vec a+\vec b$$ in the ratio $$3:1$$ is
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$$\dfrac{3\vec a-2\vec b}{2}$$
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$$\dfrac{7\vec a-8\vec b}{4}$$
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$$\dfrac{3\vec a}{4}$$
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$$\dfrac{5\vec a}{4}$$
Explanation
Let the position vector of the point $$R$$ divides the join of points $$2\vec a-3\vec b$$ and $$\vec a+\vec b$$.
$$\therefore $$ Position vector $$R=\dfrac{3( \vec a+\vec b)+1( 2\vec a-3\vec b)}{3+1}$$
Since, the position vector of a point $$R$$ divides the line segment joining the points $$P$$ and $$Q$$, whose position vectors are $$\vec p$$ and $$\vec q$$ in the ration $$m:n$$ internally, is given by $$\dfrac{m\vec q +n\vec p}{m+n}$$
$$\therefore R=\dfrac{5\vec a}{4}$$
If $$\left| \bar { a } \right| =2,\ \left| \bar { b } \right| = 3, \left| \bar { c } \right| =4$$ then $$\left[ \begin{matrix} \bar { a } +\bar { b } & \bar { b } +\bar { c } & \bar { c } -\bar { a } \end{matrix} \right] $$ is
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$$24$$
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$$-24$$
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$$0$$
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$$448$$
If $$\left| \bar { a } \right| =3,\ \left| \bar { b } \right| =4,$$ then the value of $$\lambda$$ for which $$\bar { a }+\lambda \bar { b },$$ is
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0%
$$\dfrac{9}{16}$$
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$$\dfrac{3}{4}$$
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$$\dfrac{3}{2}$$
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$$\dfrac{4}{3}$$
Let $$\bar { p }$$ and $$\bar { q }$$ be the position vectors of $$P$$
and $$Q$$ respectively, with respect to $$O$$ and $$\left| \bar { p }
\right| =p,\ \left| \bar { q } \right| =q.$$ The points $$R$$ and $$S
$$ divide $$PQ$$ internally and externally in the ratio $$2:3$$
respectively. If $$OR$$ and $$OS$$ are perpendicular; then
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$$9p^2=4q^2$$
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$$4p^2=9q^2$$
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$$9p=4q$$
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$$4p=9q$$
The value of $$\hat { i }. (\hat { j } \times \hat { k }) + \hat { j }. (\hat {
i } \times \hat { k })+\hat { k }. (\hat { i } \times \hat { j })$$ is
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0%
$$0$$
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$$-1$$
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$$1$$
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$$3$$
If $$ \overrightarrow {a} $$ is non zero vector of magnitude 'a ' and $$ \lambda $$ a nonzero scalar then $$ \lambda \overrightarrow {a} $$ is unit vector
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$$ \lambda =1 $$
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$$ \lambda = -1 $$
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$$ a = | \lambda| $$
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$$ a = 1 / | \lambda | $$
Explanation
$$|\lambda\bar{a}| \space is \space a \space unit \space vector$$
$$So \space its \space magnitude \space is \space 1$$
$$ |\lambda \bar{a}|=|\lambda||\bar{a}|=|\lambda|a=1 $$
$$ \Rightarrow a=\frac{1}{|\lambda|} $$
$$Hence\space answer\space is\space (D)$$
In triangle ABC , which of the following is not true.
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$$ \bar {AB} + \bar {BC} + \bar {CA} = \bar {0} $$
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$$ \bar {AB} + \bar {BC} - \bar {AC} = \bar {0} $$
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$$ \bar {AB} + \bar {BC} - \bar {CA} = \bar {0} $$
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$$ \bar {AB} - \bar {CB} + \bar {CA} = \bar {0} $$
Explanation
$$\textbf{Step -1: Apply triangle law of vector addition and check which option is not true.}$$
$$\text{On applying the triangle law of addition in the given triangle, we have}$$
$$\overline {AB}+\overline{BC}=\overline{AC}\ldots(i)$$
$$\overline {AB}+\overline{BC}=-\overline{CA}$$ $$\mathbf{(\because \overline{AC}=-\overline{CA})}$$
$$\overline{AB}+\overline{BC}+\overline{CA}=0\ldots(ii)$$
$$\therefore \text{Option A is true.}$$
$$\text{Equation }(i)\text{ can be writen as, }$$
$$\overline{AB}+\overline{BC}-\overline{AC}=0$$
$$\therefore \text{Option B is true.}$$
$$\text{Equation }(ii)\text{ can be written as, }$$
$$\overline{AB}-\overline{CB}+\overline{CA}=0$$ $$\mathbf{(\because \overline{BC}=-\overline{CB})}$$
$$\therefore \text{Option D is true.}$$
$$\text{Hence, option C is not true.}$$
$$\textbf{Therefore, Option C is correct.}$$
The value of $$ \hat {i} .( \hat {j} \times \hat {k}) + \hat {j} . ( \hat {i} \times \hat {k}) + \hat {k} .( \hat {i} \times \hat {j}) $$
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$$ 0 $$
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$$ -1 $$
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$$ 1 $$
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$$ 3 $$
Explanation
$$ \overrightarrow {i} . ( \overrightarrow { j} \times \overrightarrow {k} ) + \overrightarrow {j} .( \overrightarrow {i} \times \overrightarrow {k}) + \overrightarrow {k} ( \overrightarrow {i} \times \overrightarrow {j}) $$
$$ \overrightarrow {i} . \overrightarrow {i'} + \overrightarrow {j} .( - \overrightarrow {j}) + \overrightarrow {k} . \overrightarrow {k} = | i|^2 - | j|^2 + |k|^2 $$
$$ = 1 - 1 +1 = +1 $$
hence answer is ( C).
Three vectors of magnitudes $$a,\ 2a,3a$$ meeting a point and three directions are along the diagonals of three adjacent faces of a cube. The magnitude of their resultant is
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$$3a$$
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$$5a$$
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$$2a$$
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$$4a$$
Explanation
Solution:
Let the vectors of magnitudes $$a,2a,3a$$ act along $$OP,OQ,OR$$ respectively.Then vectors are $$OP,OQ,OR$$ are
$$a\left(\cfrac{\vec i+\vec j}{\sqrt2}\right),$$
$$2a\left(\cfrac{\vec j+\vec k}{\sqrt2}\right),$$
$$3a\left(\cfrac{\vec k+\vec i}{\sqrt2}\right)$$ respectively.
Their resultant say $$R$$ is given by
$$\vec R =$$
$$a\left(\cfrac{\vec i+\vec j}{\sqrt2}\right)+$$
$$2a\left(\cfrac{\vec j+\vec k}{\sqrt2}\right)+$$
$$3a\left(\cfrac{\vec k+\vec i}{\sqrt2}\right)$$
$$=\cfrac{a}{\sqrt2}(4\vec i+3\vec j+5\vec k)$$
$$\therefore |\vec R|=\sqrt{\cfrac{a^2}2(16+9+25)}=5a$$
Hence, B is the correct answer.
If $$\vec {x}$$ is a vector whose initial point divides the line joining $$5\hat{i}$$, and $$ 5\hat{j}$$ in the ratio $$\lambda :1$$ and the terminal point is the origin. Also given $$\left | \vec {x} \right |\leq \sqrt{37}$$, then $$\lambda $$ belongs to
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$$\left [ -\dfrac{1}{6} ,\dfrac{1}{6}\right ]$$
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$$(-\infty ,-6)\cup \left ( -\dfrac{1}{6} ,\infty \right )$$
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$$(-\infty ,-8)$$
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$$(1,\infty )$$
Explanation
From the condition given for $$\vec x$$ we have
$$\displaystyle \frac{|\lambda(5j)+5i|}{\lambda+1}\leq\sqrt{37}$$
$$25(\lambda^{2}+1)\leq 37(\lambda+1)^{2}$$
$$\Rightarrow(6\lambda+1)(\lambda+6)\geq 0$$
$$\displaystyle \lambda\in(-\infty, -6)\mathrm\cup (-\frac{1}{6}, \infty)$$
A scooterist follows a track on a ground that turns to his left by an angle 60$$^{0}$$ after every 400 m. Starting from the given point displacement of the scooterist at the third turn and eighth turn are :
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$$800\mathrm{ m}; 0\mathrm{ m}$$
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$$800\mathrm{m},\ 800\sqrt{3}\mathrm{m}$$
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$$800\mathrm{m};400\sqrt{3}\mathrm{m}$$
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$$800; 800\sqrt{3}\mathrm{m}$$
Explanation
Let $$\bar { { x }_{ 1 } } ,\bar { { x }_{ 2 } } ,\bar { { x }_{ 3 } } ,.....\bar { { x }_{ 8 } } $$ be the positions of scooterist after $${ 1 }^{ st },{ 2 }^{ nd },{ 3 }^{ rd }......,{ 8 }^{ th }$$ turn respectively.
Thus its motion can be visualized as hexagon inscribed in circle of radius $$400m$$.
Let $$\bar { { x }_{ 0 } } $$ be its initial position.
$$\therefore $$ After $${ 3 }^{ rd }$$ turn
$${ x }_{ 3 } -{ x }_{ 0 }=800m$$
After $${ 8 }^{ th }$$ turn
$${ x }_{ 0 } - { x }_{ 2 } = 800\cos30^{0} = 400\sqrt 3m$$
The vectors $$\overrightarrow{AB} = 3\hat{i} + 4\hat{k}$$ and $$\overrightarrow{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$$ are the sides of a triangle $$ABC$$, then the length of the median through $$A$$ is:
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$$\sqrt{72}$$
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$$\sqrt{33}$$
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$$\sqrt{45}$$
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$$\sqrt{18}$$
Explanation
Take A as the origin(0,0)
so $$\vec {AB}= 3\hat i + 4\hat k$$ and $$AC = 5\hat i – 2\hat j + 4\hat k$$ will become position vector.
So coordinate of B is $$(3,0,4)$$ and C is$$(5,-2,4)$$.
The mid point of BC can be easily found by midpoint formula of two points as $$D(4,-1,4).$$
so length of median through $$A$$ is $$AD$$$$=\sqrt{(4-0)^2+(-1-0)^2+(4-0)^2}$$
$$=\sqrt{33}$$
$$\mathrm{l}\mathrm{n}$$ a triangle O$$\mathrm{A}\mathrm{B},\ \mathrm{E}$$ is the mid-point of $$\mathrm{O}\mathrm{B}$$ and $$\mathrm{D}$$ is a point on $$\mathrm{A}\mathrm{B}$$ such that $$\mathrm{A}\mathrm{D}$$: $$\mathrm{D}\mathrm{B}=2: 1$$. lf $$\mathrm{O}\mathrm{D}$$ and $$\mathrm{A}\mathrm{E}$$ interesect at $$\mathrm{P}$$, then the ratio $$\displaystyle\frac{OP}{PD}$$ is
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$$1:2$$
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$$3:2$$
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$$8:3$$
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$$4:3$$
Explanation
$$\Delta OAB$$ let $$O=(0,0),A=(x_1,y_1),B=(x_2,y_2)$$
Then, $$E=(\dfrac{x_2}{2},\dfrac{y_2}{2})$$,$$D=(\dfrac{2x_2+x_1}{3},\dfrac{2y_2+y_1}{3})$$
Equation of OD is: $$y-0=\dfrac{2y_2+y_1}{2x_2+x_1}(x-0)$$
$$y=\dfrac{2y_2+y_1}{2x_2+x_1}x$$........(1)
Equation of AE is: $$(y-y_1)=\dfrac{y_2-2y_1}{x_2-2x_1}(x-x_1)$$
From eq(1)
$$(\dfrac{2y_2+y_1}{2x_2+x_1})(x)-y_1=\dfrac{y_2-2y_1}{x_2-2x_1}(x-x_1)$$
so $$ x=\dfrac{(2x_2+x_1)}{5} =\dfrac{3(\dfrac{2x_2+x_1}{3})+2(0)}{(3+2)}$$
so ratio $$OP:PD=3:2$$
In a quadrilateral $$PQRS,\ \vec{PQ}=\vec{a}, \vec{QR}=\vec{b}, \vec{SP}=\vec{a} - \vec{b}.\ M$$ is the mid-point of $$QR$$ and $$X$$ is a point on $$SM$$ such that $$\vec{SX}=\dfrac{4}{5}\vec{SM}$$, then $$\vec{PX}$$ is
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$$\dfrac{1}{5}\vec{PR}$$
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$$\dfrac{3}{5}\vec{PR}$$
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$$\dfrac{2}{5}\vec{PR}$$
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None of these
Explanation
Choose $$P$$ as the origin of reference.
$$\vec{PR}=\vec{a}+\vec{b}$$,
$$\vec{PS}=-(\vec{a}-\vec{b})=\vec{b}-\vec{a}$$
$$\vec{PM}=\vec{a}+\dfrac{\vec{b}}{2}$$
Given :
$$\vec{SX}=\dfrac{4}{5}\vec{SM}$$
$$\vec{PX}-(\vec{b}-\vec{a})=\dfrac{4}{5}\left \{ \vec{a} +\dfrac{\vec{b}}{2}-(\vec{b}-\vec{a})\right \}$$
$$\Rightarrow \vec{PX}=\dfrac{3}{5}(\vec{a}+\vec{b})$$
$$=\dfrac{3}{5}\vec{PR}$$
Hence, option $$B.$$
The position vectors of $$A$$ and $$B$$ are $$2\hat{i}+2\hat{j}+\hat{k}$$ and $$2\hat{i}+4\hat{j}+4\hat{k}.$$ The length of the internal bisector of $$\angle BOA$$ of the triangle $$AOB$$ is
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$$\displaystyle \sqrt { \dfrac { 136 }{ 9 } } $$
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$$\displaystyle \sqrt { \dfrac { 139 }{ 9 } } $$
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$$\displaystyle \dfrac { 20 }{ 3 } $$
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$$\displaystyle \sqrt { \dfrac { 217 }{ 9 } } $$
Explanation
Here $$OA=3,OB=6$$
$$\therefore$$ Internal bisector $$OD$$ divides $$AB$$ in the ratio $$1:2$$
$$\Rightarrow$$ Position vector of $$D$$ is $$\displaystyle \left( 2,\frac { 8 }{ 3 } ,2 \right) $$
$$\displaystyle \therefore \left| OD \right| =\sqrt { 4+\dfrac { 64 }{ 9 } +4 } =\sqrt { \dfrac { 136 }{ 9 } } $$
$$ABCD$$ is a quadrilateral, $$E$$ is the point of intersection of the line joining the midpoints of the opposite sides. If $$O$$ is any point and $$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{x OE},$$ then $$x$$ is equal to
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$$3$$
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$$9$$
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$$7$$
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$$4$$
Explanation
Let $$\vec{OA} = \vec{a}, \vec{OB} = \vec{b}, \vec{OC} = \vec{c}$$ and $$\vec{OD} = \vec{d}$$.
Therefore,
$$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = \vec{a} + \vec{b} + \vec{c} + \vec{d}$$.
The midpoint $$P$$ of $$AB$$, is $$\displaystyle \dfrac{\vec{a} + \vec{b}}{2}$$.
The position vector of the midpoint $$Q$$ of $${CD}$$, is $$\displaystyle \dfrac{\vec{c} + \vec{d}}{2}$$.
Therefore, the position vector of the midpoint of $${PQ} $$ is $$\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$$.
Similarly, the position vector of the midpoint of $$RS$$ is $$\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$$, i.e., $$\vec{OE} =\displaystyle \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4} \Rightarrow x = 4$$
Hence, option '$$D$$' is correct.
If $$\overrightarrow{b}$$ is a vector whose initial point divides the join of $$5\widehat{i}$$ and $$5\widehat{j}$$ in the ratio $$k : 1$$ and whose terminal point is the origin and $$|\vec b| \leq \sqrt{37}$$, then $$k$$ lies in the interval
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$$\left[-6, -\dfrac{1}{6}\right]$$
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$$\left(- \infty, -6 \right] \cup \left[-\dfrac{1}{6}, \infty \right) $$
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$$ \left[0, 6 \right] $$
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None of these
Explanation
The point that divides 5$$\widehat{i}$$ and 5$$\widehat{j}$$ in the ratio of $$k : 1$$ is
$$\displaystyle \dfrac{(5 \widehat{j}) k + (5\widehat{i})1}{k + 1}$$
$$\therefore \displaystyle \vec b = \frac{5 \widehat{i} + 5 k \widehat{j}}{k + 1}$$
Also, $$|\vec b| \leq \sqrt{37}$$
$$\Rightarrow \displaystyle \dfrac{1}{k+1} \sqrt{25 + 25 k^2} \leq \sqrt{37}$$
or $$5\sqrt{1 + k^2} \leq \sqrt{37} (k + 1)$$
Squaring both sides, we get
$$25 (1 + k^2) \leq 37 (k^2 + 2k +1)$$
or $$6k^2 + 37k + 6 \geq 0$$ or $$(6k + 1) (k + 6) \geq 0$$
$$k \in (-\infty, - 6] \cup \left [ - \dfrac{1}{6}, \infty \right )$$
If $$\vec{a},\vec{b},\vec{c}$$ are three non-zero vectors such that $$\vec{a}\times \vec{b}=\vec{c}$$ and $$\vec{b}\times \vec{c}=\vec{a}$$ , then choose the
incorrect
option(s)
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$$\vec{a}. \vec{b}=\vec{b}. \vec{c}=\vec{c} . \vec{a}=0$$
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$$\left | \vec{b} \right |=\left | \vec{c} \right |$$
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$$\vec{a}$$ is a unit vector
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$$\vec{c}$$ is a unit vector
Explanation
$$\vec{a}\times \vec{b}=\vec{c}\Rightarrow \vec{a}.\vec{c}=0=\vec{b}.\vec{c}$$
$$\vec{b}\times \vec{c}=\vec{a}\Rightarrow \vec{a}.\vec{b}=0=\vec{a}.\vec{c}$$
$$\Rightarrow \vec{a},\vec{b},\vec{c} $$ are mutually pependicular vectors.
$$\therefore \vec{a}\times \vec{b}=\vec{c}\Rightarrow \left | \vec{a} \right |\left | \vec{b} \right |=\left | \vec{c} \right |$$
$$\vec{b}\times \vec{c}=\vec{a}\Rightarrow \left | \vec{b} \right |\left | \vec{c} \right |=\left | \vec{a} \right |$$
$$\therefore \left | \vec{b} \right |^{2}\left | \vec{c} \right |=\left | \vec{c} \right |\Rightarrow \left | \vec{b} \right |=1$$
$$\therefore \left | \vec{a} \right |=\left | \vec{c} \right |$$
Hence, options B , C and D are incorrect.
If the vectors $$\hat i - \hat j, \hat j + \hat k$$ and $$\vec a$$ form a triangle, then $$\vec a$$ may be
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$$- \hat i - \hat k$$
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$$\hat i - 2 \hat j - \hat k$$
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$$2 \hat i + \hat j + \hat k$$
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$$\hat i + \hat k$$
Explanation
If $$\vec{p}, \vec{q}, \vec{r}$$ form a triangle, only three cases exist :
1. $$\vec{p} + \vec{q} = \vec{r}$$
2. $$\vec{q} + \vec{r} = \vec{p}$$
3. $$\vec{r} + \vec{p} = \vec{q}$$
Similarly, 1. $$\vec{a} = \vec{i} + \vec{k}$$
2. $$\vec{a} = (\vec{j} + \vec{k}) - (\vec{i} - \vec{j}) = - \vec{i} + 2\vec{j} + \vec{k}$$
3. $$\vec{a} = -(\vec{j} + \vec{k}) + (\vec{i} - \vec{j}) = \vec{i} - 2\vec{j} - \vec{k}$$
Vectors $$\vec a = \hat i + 2 \hat j + 3 \hat k, \vec b = 2 \hat i - \hat j + \hat k$$ and $$\vec c = 3 \hat i + \hat j + 4 \hat k$$ are so placed that the end point of one vector is the starting point of the next vector, then the vectors are
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Not coplanar
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Coplanar but cannot form a triangle
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Coplanar and form a triangle
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Coplanar and can form a right-angled triangle
Explanation
Given $$\vec{AB}=\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$
$$\vec{BC}=\vec{b}=2\hat{i}-\hat{j}+\hat{k}$$
$$\vec{AC}=\vec{c}=3\hat{i}+\hat{j}+4\hat{k}$$
To check coplanar $$[\vec{a}\vec{b}\vec{c}]=0$$
\begin{vmatrix}1&2&3\\ 2&-1&1\\ 3&1&4 \end{vmatrix}=0
$$1(-5)-2(5)+3(5)=0$$
It is coplanar
Now to check a triangular
$$\vec{AC}=\vec{AB}+\vec{BC}$$
$$\hat{i}+2\hat{j}+3\hat{k}+2\hat{i}-\hat{j}+\hat{k}=\vec{AC}$$
$$3\hat{i}+\hat{j}+4\hat{k}=3\hat{i}+\hat{j}+4\hat{k}=\vec{AC}$$
So condition is followed it is triangular
$$ABCD$$ a parallelogram, $$A_1$$ and $$B_1$$ are the midpoints of sides $$BC$$ and $$CD$$, respectively. If $$\vec{AA_1} + \vec{AB_1} = \lambda \vec{AC}$$, then $$\lambda$$ is equal to
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$$\displaystyle \dfrac{1}{2}$$
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$$1$$
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$$\displaystyle \dfrac{3}{2}$$
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$$2$$
Explanation
Let P.V. of $$A, B$$ and $$D$$ be $$\vec 0, \vec b,$$ and $$\vec d$$, respectively.
Then P.V. of $$C, \vec c = \vec b + \vec d$$
Also, P.V of $$A_1 = \vec b + \displaystyle \dfrac{\vec d}{2}$$
and P.V. of $$B_1 = \vec d + \displaystyle \dfrac{\vec b}{2}$$
$$\Rightarrow \vec{AA_1} + \vec{AB_1} = \displaystyle \dfrac{3}{2} (\vec b + \vec d) = \dfrac{3}{2} \vec{AC}$$
$$L_{1}and L_{2}$$ are two lines whose vector equations are
$$L_{1}:\vec{r}=\lambda \left ( (\cos \theta+\sqrt{3})\hat{i}+(\sqrt{2}\sin\theta)\hat{j}+(\cos \theta-\sqrt{3})\hat{k} \right )$$
$$L_{2}:\vec{r}=\mu \left ( a \hat{i}+b \hat{j}+c\hat{k} \right ),$$
Where $$\lambda\ and\ \mu $$ are scalars and $$\alpha$$ is the acute angle
between $$L_{1}\ and\ L_{2}$$ . If the angle '$$\alpha$$'is independent
of $$\theta $$ then the value of '$$\alpha$$ ' is
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$$\dfrac{\pi}{6}$$
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{2}$$
Explanation
we know that $$cos\alpha=\dfrac{L_1.L_2}{|L_1|.|L_2|}$$ where $$\alpha$$ is the angle between the vectors $$L_1,L_2$$.
$$\implies cos\alpha=\dfrac{\lambda\mu(a(cos\theta+\sqrt 3)+b(\sqrt 2sin\theta)+c(cos\theta-\sqrt 3))}{\lambda\mu(\sqrt{a^2+b^2+c^2}).(\sqrt{cos^2\theta+3+2\sqrt 3cos\theta+2sin^2\theta+cos^2\theta+3-2\sqrt 3cos\theta})}$$
$$\implies cos\alpha=\dfrac{(a+c)cos\theta+\sqrt 3(a-c)+b\sqrt 2sin\theta}{8(\sqrt{a^2+b^2+c^2})}$$
For $$\alpha$$ to be independent of $$\theta$$
$$a+c=0$$ ..........(1)
and $$b=0$$ ........(2)
Therefore, $$cos\alpha=\dfrac{\sqrt 3(a-c)}{\sqrt 8.(\sqrt{a^2+c^2})}$$
$$\implies cos\alpha=\dfrac{\sqrt 3(2a)}{(\sqrt{8\times2a^2})}$$
$$\implies cos\alpha=\dfrac{\sqrt 3}{2}$$
$$\implies \alpha=\dfrac{\pi}{6}$$
$$\overrightarrow{AR}$$ is
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$$\dfrac{1}{5}(2\vec {b}+\vec {c})$$
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$$\dfrac{1}{6}(2\vec {b}+\vec {c})$$
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$$\dfrac{1}{7}(\vec {b}+2\vec {c})$$
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$$\dfrac{1}{7}(2\vec {b}+\vec {c})$$
Explanation
Choosing $$A$$ as the origin of reference , we find $$\bar{b}$$ and $$\, \bar{c}$$ are the position vectors of $$B$$ and $$C$$ respectively.
Position vector of $$P$$ is $$\displaystyle \dfrac{\bar{b}}{3}$$
Position vector of $$Q$$ is $$\displaystyle \dfrac{2\bar{b}+\bar{c}}{3}$$
$$\therefore $$ Lets take $$R$$ divides $$AQ$$ in the ratio $$(1-t):t
$$
$$\displaystyle \overline{r}=(1-t)\overline{o}+t\left(\dfrac{2\overline{b}+\overline{c}}{3}\right)$$ where $$t$$ is a scalar
$$\overrightarrow { r } =t\left( \dfrac { 2b+c }{ 3 } \right) $$
Lets take $$R$$ divides $$CP$$ in the ratio $$s:1-s $$ is
$$\overline{r}=(1-s)\overline{c}+s \overline{\dfrac{b}{3}}$$, where $$s$$ is a scalar
If $$AQ$$ and $$CP$$ are to intersect
$$\displaystyle t\left(\dfrac{2\overline{b}+\overline{c}}{3}\right)=(1-s)\overline{c}+s\overline{ \dfrac{b}{3}}$$ for same $$t$$ and $$s$$
Equating like components
$$\displaystyle \dfrac{2t}{3}=\dfrac{s}{3}$$ and $$\displaystyle \dfrac{t}{3}=1-s$$
On solving we get,
$$\displaystyle t=\dfrac{3}{7}$$ and $$\displaystyle s=\dfrac{6}{7}$$
$$\therefore $$ $$\displaystyle \overline{AR}=\dfrac{3}{7}\left(\dfrac{2\overline{b}+\overline{c}}{3}\right)=\dfrac{1}{7}(2\overline{b}+\overline{c})$$
In a parallelogram $$OABC,$$ vectors $$\vec{a}, \vec{b}, \vec{c}$$ are, respectively, the position vectors of vertices $$A, B, C$$ with reference to $$O$$ as origin. A point $$E$$ is taken on the side $$BC$$ which divides it in the ratio of $$2 : 1$$. Also, the line segment $$AE$$ intersects the line bisecting the angle $$\angle$$AOC internally at point $$P$$. If $$CP$$ when extended meets $$AB$$ in point $$F,$$ then t
he position vector of point $$P$$ is
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$$\displaystyle \dfrac{|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$
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$$\displaystyle \dfrac{3|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$
0%
$$\displaystyle \dfrac{2|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$
0%
None of these
Explanation
Let the position vector of $$A$$ and $$C$$ be $$\vec{a}$$ and $$\vec{c}$$ respectively. Therefore,
Position vector of $$B = \vec{b} = \vec{a} + \vec{c}$$...........................................................$$(i)$$
Also Position vector of $$E = \displaystyle \dfrac{\vec{b} + 2 \vec{c}}{3} = \dfrac{\vec{a} + 3 \vec{c}}{3}$$ ............................................................$$(ii)$$
Now point $$P$$ lies on angle bisector of $$\angle AOC.$$ Thus,
Position vector of point $$P =\displaystyle \lambda \left ( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right )$$............................................................................$$(iii)$$
Also let $$P$$ divides $$EA$$ in ration $$\mu : 1$$. Therefore,
Position vector of $$P$$
$$\displaystyle = \dfrac{\mu \vec{a} + \dfrac{\vec{a} + 3 \vec{c}}{3}}{\mu + 1} = \dfrac{(3 \mu + 1) \vec{a} + 3 \vec{c}}{3 (\mu + 1)}$$................................................................................$$(iv)$$
Comparing $$(iii)$$ and $$(iv),$$ we get
$$\lambda \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right ) = \dfrac{ (3 \mu + 1) \vec{a} + 3 \vec{c}}{3 (\mu + 1)}$$
$$\Rightarrow \displaystyle \dfrac{\lambda}{|\vec{a}|} = \dfrac{3 \mu + 1}{3 (\mu + 1)}$$ and $$\displaystyle \dfrac{\lambda}{|\vec{c}|} = \dfrac{1}{\mu + 1}$$
$$\Rightarrow \displaystyle \dfrac{3 |\vec{c}| - |\vec{a}|}{3 |\vec{a}|} = \mu$$
$$\displaystyle \Rightarrow \dfrac{\lambda}{|\vec{c}|} = \dfrac{1}{\dfrac{3 |\vec{c}| - \vec{a}}{3 |\vec{a}|}+1}$$
$$\Rightarrow \displaystyle \lambda = \dfrac{3 |\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|}$$
Hence, position vector of $$P$$ is $$\displaystyle \dfrac{3 |\vec{a}| |\vec{c}|}{ 3 |\vec{c}| + 2 |\vec{c}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$$
Let $$F$$ divides $$AB$$ in ratio $$t : 1,$$ then position vector of $$F$$ is $$\displaystyle \dfrac{t \vec{b} + \vec{a}}{ t + 1} $$
The ratio $$\displaystyle \dfrac{OX}{XC}$$ is
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$$\dfrac{3}{4}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{2}{5}$$
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$$\dfrac{1}{2}$$
Explanation
Let the position vectors of $$A, B, C$$ and $$D$$ be $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$, respectively.
Then, $$OA : CB = 2 : 1$$
$$\Rightarrow \vec{OA} = 2 \vec{CB}$$
$$\Rightarrow \vec{a} = 2 (\vec{b} - \vec{c})$$ ....$$(i)$$
and $$OD : AB = 1 : 3$$
$$3\vec{OD} = \vec{AB}$$
$$\Rightarrow 3\vec{d} = (\vec{b} - \vec{a}) = \vec{b} - 2 (\vec{b} - \vec{c})$$ [Using $$(i)$$]
$$= -\vec{b}+ 2\vec{c}$$ .....$$(ii)$$
Let $$OX : XC = \lambda : 1$$ and $$AX : XD =\mu : 1$$
Now,
$$X$$ divides $$OC$$ in the ratio $$\lambda : 1$$.
Therefore,
P.V of $$X = \displaystyle \frac{\lambda \vec{c}}{\lambda + 1}$$ ....$$(iii)$$
$$X$$ also divides $$AD$$ in the ratio $$\mu : 1$$.
Therefore,
P.V. of $$\displaystyle X = \dfrac{\mu \vec{d} + \vec{a}}{\mu + 1}$$ ...$$(iv)$$
From $$(iii)$$ and $$(iv)$$, we get
$$\displaystyle \dfrac{\lambda \vec{c}}{\lambda + 1} = \dfrac{\mu \vec{d} + \vec{a}}{\mu + 1}$$
or $$\displaystyle \left ( \frac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \frac{\mu}{\mu + 1} \right )\vec{d} + \left ( \frac{1}{\mu + 1} \right ) \vec{a}$$
or $$\displaystyle \left (\dfrac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \dfrac{\mu}{\mu + 1} \right ) \left ( \dfrac{-\vec{b} + 2 \vec{c}}{3} \right ) + \left ( \dfrac{1}{\mu + 1} \right ) 2 \left ( \vec{b} - \vec{c} \right )$$ ....(using $$(i)$$ and $$(ii)$$)
or $$\displaystyle \left ( \frac{\lambda}{\lambda + 1} \right ) \vec{c}= \left ( \frac{6 - \mu}{3 (\mu + 1)} \right ) \vec{b} + \left ( \frac{2 \mu}{3 (\mu + 1)} - \frac{2}{\mu + 1} \right ) \vec{c}$$
or $$\displaystyle \left ( \dfrac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \dfrac{6 - \mu}{3 (\mu + 1)} \right ) \vec{b} + \left ( \dfrac{2\mu - 6}{3 (\mu + 1)} \right ) \vec{c}$$
or $$\displaystyle \left ( \frac{6 - \mu}{3 (\mu + 1)} \right )\vec{b} + \left ( \frac{2 \mu - 6}{3 (\mu + 1)} - \frac{\lambda}{\lambda + 1} \right ) \vec{c} = \vec{0}$$
or $$\displaystyle \dfrac{6 - \mu}{3 (\mu + 1)} = 0$$ and $$\displaystyle \dfrac{2 \mu - 6}{3 (\mu + 1)} - \dfrac{\lambda}{\lambda + 1} = 0$$
(as $$\vec{b}$$ and $$\vec{c}$$ are non-collinear)
or $$\displaystyle \mu = 6, \lambda = \dfrac{2}{5}$$
Hence, $$ OX : XC = 2 : 5$$
The projection of the line joining the points $$(3, 4, 5)$$ and $$(4, 6, 3)$$ on the line joining the points $$(-1, 2, 4)$$ and $$(1, 0, 5)$$ is
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$$\dfrac{4}{3}$$
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$$\dfrac{2}{3}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{2}$$
Explanation
Let $$AB=(4-3)\hat i+(6-4)\hat j+(3-5)\hat k=\hat i+2\hat j-2\hat k$$ and $$PQ=(1+1)\hat i+(0-2)\hat j+(5-4)\hat k=2\hat i-2\hat j+\hat k$$
Hence projection of $$AB$$ on $$PQ$$ is $$=\left |\cfrac{AB\cdot PQ}{|AB||PQ|}\right |$$
$$=\left |\cfrac{1(2)+2(-2)-2(1)}{\sqrt{3}\sqrt{3}}\right |$$
$$=\cfrac{4}{3}$$
A parallelogram is constructed on the vectors $$\bar{\alpha }$$ and $$\bar{\beta }$$. A vector which coincides with the altitude of the parallelogram and perpendicular to the side $$\bar{\alpha }$$ expressed in terms of the vectors $$\bar{\alpha }$$ and $$\bar{\beta }$$ is
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$$\displaystyle \bar{\beta }+\dfrac{\bar{\beta }-\bar{\alpha }}{\left ( \bar{\alpha } \right )^{2}}\bar{\alpha }$$
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$$\displaystyle \dfrac{\left ( \bar{\alpha }\times \bar{\beta } \right )\times \bar{\alpha }}{\left ( \bar{\alpha } \right )^{2}}$$
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$$\displaystyle \dfrac{\bar{\beta }\cdot \bar{\alpha }}{\left ( \alpha \right )^{2}}\bar{\alpha }+\bar{\beta }$$
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$$\displaystyle \left | \bar{\beta } \right |\dfrac{\bar{\alpha }\times \left ( \bar{\alpha }\times \bar{\beta } \right )}{\left ( \alpha \right )^{2}}$$
Explanation
$$(\alpha\times\beta)$$ will give us the perpendicular vector or the normal to the plane where the parallelogram is lying.
Now
$$(\alpha\times \beta)\times \alpha$$ will give the vector parallel to the altitude of the parallelogram which is perpendicular to side $$\bar {\alpha}$$.
Thus the unit vector of the altitude of the given parallelogram perpendicular to $$\bar{\alpha}$$ will be
$$=\dfrac{(\alpha\times\beta)\times\alpha}{|\alpha|^{2}}$$
$$\vec{a} = 2 \widehat{i} - \widehat{j} + \widehat{k}, \vec{b} = \widehat{i} + 2\widehat{j} - \widehat{k} $$ and $$ \vec{c} = \widehat{i} + \widehat{j} - 2 \widehat{k}$$. A vector coplanar with $$\vec{b}$$ and $$\vec{c}$$ whose projection on $$\vec{a}$$ is magnitude $$\displaystyle \sqrt{\dfrac{2}{3}} $$ is
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$$2 \widehat{i} + 3 \widehat{j} - 3 \widehat{k}$$
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$$- 2 \widehat{i} - \widehat{j} + 5 \widehat{k}$$
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$$2 \widehat{i} + 3 \widehat{j} + 3 \widehat{k}$$
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$$2 \widehat{i} + \widehat{j} + 5 \widehat{k}$$
Explanation
Let the required vector be $$\vec{r}$$, then
$$\vec{r} = x_1 \vec{b} + x_2 \vec{c}$$ and $$\vec{r} \cdot \vec{a} = \displaystyle \sqrt{\dfrac{2}{3}} (|\vec{a}|) = 2$$
Now, $$\vec{r} \cdot \vec{a} = x_1 \vec{a} \cdot \vec{b} + x_2 \vec{a} \cdot \vec{c}$$
$$\Rightarrow 2 = x_1 (2 - 2 - 1) + x_2 (2 - 1 - 2)$$
$$\Rightarrow x_1 + x_2 = -2$$
$$\Rightarrow \vec{r} = x_1 (\widehat{i} + 2 \widehat{j} - \widehat{k}) + x_2 (\widehat{i} + \widehat{j} - 2 \widehat{k})$$
$$= \widehat{i} (x_1 + x_2) + \widehat{j} (2x_1 + x_2) - \widehat{k} (2x_2 + x_1)$$
$$= - 2\widehat{i} + \widehat{j} (x_1 - 2) - \widehat{k} (-4 - x_1)$$,
where $$x_1 \in R$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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