MCQ Questions for 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

0%
isosceles
0%
equilateral
0%
right-angled
0%
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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$\dfrac{1}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times(\vec{a} \times \vec{b})]$
0%
$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$
0%
$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$
0%
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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$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b} \times(\vec{a} \times \vec{b})]$
0%
$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$
0%
$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$
0%
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

0%
$\dfrac{\vec{a} \times \vec{b}}{\gamma}$
0%
$\vec{a}+\dfrac{\vec{a} \times \vec{b}}{\gamma}$
0%
$\vec{a}+\vec{b}+\dfrac{\vec{a} \times \vec{b}}{\gamma}$
0%
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

0%
$\vec{P}$
0%
$-\vec{P}$
0%
$2 \vec{B}$
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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

0%
$2 \mid \vec p \cdot \vec q \mid$
0%
$(1/2) \mid \vec p \cdot \vec q \mid$
0%
$\mid \vec p \times \vec q \mid$
0%
$\mid \vec p \cdot \vec q \mid$

Explanation

$\vec a + \vec b = \mu \vec p,$

$\vec b \cdot \vec q = 0$ ,

$(\vec b)^2 = 1,$

$\because \vec a + \vec b = \mu \vec p$

$\Rightarrow (\vec a + \vec b) \times \vec a = \mu \vec p \times \vec a, \vec b \times \vec a = \mu \vec p \times \vec a \Rightarrow \vec q \times (\vec b \times \vec a) = \mu \vec q \times (\vec p \times \vec q)$

$\Rightarrow (\vec q \cdot \vec a) \vec b - (\vec q \cdot \vec b) \vec a = \mu \vec q \times (\vec p \times \vec a) \Rightarrow (\vec q \cdot \vec a) \vec b = \mu \vec q \times (\vec p \times \vec a)$

$\because \vec a + \vec b = \mu \vec p$

$\Rightarrow \vec q \cdot (\vec a + \vec b) = \mu \vec q \cdot \vec p$

$\Rightarrow \vec q \cdot \vec a + \vec q \cdot \vec b = \mu \vec p \cdot \vec q$

$\mu = \dfrac{\vec q \cdot \vec a}{\vec p \cdot \vec q}$

$\Rightarrow (\vec q \cdot \vec a) \vec b = \dfrac{\vec q \cdot \vec a}{\vec p \cdot \vec q} [(\vec q \cdot \vec a) \cdot \vec p - (\vec q \cdot \vec p)\vec a]$

$\Rightarrow \mid (\vec q \cdot \vec a)\vec p - (\vec q \cdot \vec p)\vec a \mid = \mid (\vec p \cdot \vec q) \vec b \mid = \mid (\vec p \cdot \vec q)\mid \cdot \mid \vec b\mid$

$\Rightarrow \mid (\vec q \cdot \vec a)\vec p - (\vec q \cdot \vec p) \vec a \mid = \mid \vec p \cdot \vec q \mid$

$\vec r$ 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

0%
$2 \mid \vec r \mid^2$
0%
$\mid \vec r \mid^2/2$
0%
$3 \mid \vec r \mid^2$
0%
$\mid \vec r \mid^2$

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

$\mid b \vec s\mid^2 + \mid \vec r + b \vec s \mid^2 = \mid - \vec r \mid^2 + \mid \vec r - \vec r \mid^2 = \mid \vec r \mid^2$

$\overline a$ and

$\overline b$ are adjacent sides of a rhombus, then

$\overline a.\overline b=0$ .

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True
0%
False

Explanation

$\overline a.\overline b=0$ , 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

$\vec{r}$ .

0%
$\vec{r} = (\vec{r} . \hat{a})\hat{a} + (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$
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$\vec{r} = (\vec{r} . \hat{a}) - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b}))) (\hat{a} \times \hat{b})$
0%
$\vec{r} = (\vec{r} . \hat{a})\hat{a} - (\vec{r} . \hat{b})\hat{b} + (\vec{r} . (\hat{a} \times \hat{b})) (\hat{a} \times \hat{b})$
0%
none of these

The vector with initial point

$P(2,-3,5)$ and terminal point

$Q(3,-4,7)$ is

0%
$\hat i-\hat j+2\hat k$
0%
$5\hat i-7\hat j+12\hat k$
0%
$\hat i+\hat j-2\hat k$
0%
$None\ of\ these$

Explanation

$(A)$ is the correct answer.
Required vector

$=|\vec{AB}|=(3-2) \hat i+(-4+3)\hat j+(7-5)\hat k$

$=\hat i-\hat j+2\hat k$

$\vec{X} \cdot \vec{A}=0, \vec{X} \cdot \vec{B}=0$ and

$\vec{X} \cdot \vec{C}=0$ for some non-zero vector

$\vec{X},$ then

$[\vec{A} \vec{B} \vec{C}]=0$

0%
True
0%
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

0%
$\dfrac {3\vec a+2\vec b}{3}$
0%
$\vec a$
0%
$\dfrac {5\vec a-\vec b}{3}$
0%
$\dfrac {4\vec a+\vec b}{3}$

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

0%
$\dfrac {2}{3}$
0%
$\dfrac {1}{3}$
0%
$2$
0%
$\sqrt 6$

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.

0%
True
0%
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

0%
$47$
0%
$-25$
0%
$0$
0%
$25$
0%
$50$

Line

$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$ will not meet the plane

$\overrightarrow{r} \cdot \overrightarrow{n} = q$ , if

0%
$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$
0%
$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$
0%
$\overrightarrow{b} \cdot \overrightarrow{n} = 0, \overrightarrow{a} \cdot \overrightarrow{n} \neq q$
0%
$\overrightarrow{b} \cdot \overrightarrow{n} \neq 0, \overrightarrow{a} \cdot \overrightarrow{n} = q$

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.

$\vec \alpha | =4$ and

$-3 \le \lambda \le 2$ , then the range of

$| \lambda \vec \alpha |$ is

0%
$[0, 8]$
0%
$[-12, 8]$
0%
$[0, 12]$
0%
$[8, 12]$

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,

$-3 \le \lambda \le 2$

$0 \le || \lambda | \le 3$

$\Rightarrow 0 \le 4 | \lambda | \le 12$

$| \lambda \vec a| \in [ 0, 12]$

$\vec a, \vec b, \vec c$ 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

0%
$1$
0%
$3$
0%
$-\dfrac 32$
0%
$None\ of\ these$

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

0%
$-\hat i +12 \hat j +4\hat k$
0%
$5\hat i+2\hat j-4\hat k$
0%
$-5\hat i +2\hat j+4\hat k$
0%
$\hat i+\hat j+ \hat k$

Explanation

Required vector

$=|\vec{AB}|=(-3-2) \hat i+(7-5)\hat j+(4-0)\hat k$

$=-5\hat i+2\hat j=4\hat k$

$\vec a, \vec b, \vec c$ 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

0%
$0$
0%
$1$
0%
$-19$
0%
$38$

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

0%
$\dfrac{3\vec a-2\vec b}{2}$
0%
$\dfrac{7\vec a-8\vec b}{4}$
0%
$\dfrac{3\vec a}{4}$
0%
$\dfrac{5\vec a}{4}$

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

$\dfrac{m\vec q +n\vec p}{m+n}$

$\therefore R=\dfrac{5\vec a}{4}$

$\left| \bar { a } \right| =2,\ \left| \bar { b } \right| = 3, \left| \bar { c } \right| =4$ then

$\left[ \begin{matrix} \bar { a } +\bar { b } & \bar { b } +\bar { c } & \bar { c } -\bar { a } \end{matrix} \right]$ is

0%
$24$
0%
$-24$
0%
$0$
0%
$448$

$\left| \bar { a } \right| =3,\ \left| \bar { b } \right| =4,$ then the value of

$\lambda$ for which

$\bar { a }+\lambda \bar { b },$ is

0%
$\dfrac{9}{16}$
0%
$\dfrac{3}{4}$
0%
$\dfrac{3}{2}$
0%
$\dfrac{4}{3}$

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

0%
$9p^2=4q^2$
0%
$4p^2=9q^2$
0%
$9p=4q$
0%
$4p=9q$

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

0%
$0$
0%
$-1$
0%
$1$
0%
$3$

$\overrightarrow {a}$ is non zero vector of magnitude 'a ' and

$\lambda$ a nonzero scalar then

$\lambda \overrightarrow {a}$ is unit vector

0%
$\lambda =1$
0%
$\lambda = -1$
0%
$a = | \lambda|$
0%
$a = 1 / | \lambda |$

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.

0%
$\bar {AB} + \bar {BC} + \bar {CA} = \bar {0}$
0%
$\bar {AB} + \bar {BC} - \bar {AC} = \bar {0}$
0%
$\bar {AB} + \bar {BC} - \bar {CA} = \bar {0}$
0%
$\bar {AB} - \bar {CB} + \bar {CA} = \bar {0}$

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

$\hat {i} .( \hat {j} \times \hat {k}) + \hat {j} . ( \hat {i} \times \hat {k}) + \hat {k} .( \hat {i} \times \hat {j})$

0%
$0$
0%
$-1$
0%
$1$
0%
$3$

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

0%
$3a$
0%
$5a$
0%
$2a$
0%
$4a$

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.

$\vec {x}$ 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

0%
$\left [ -\dfrac{1}{6} ,\dfrac{1}{6}\right ]$
0%
$(-\infty ,-6)\cup \left ( -\dfrac{1}{6} ,\infty \right )$
0%
$(-\infty ,-8)$
0%
$(1,\infty )$

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 :

0%
$800\mathrm{ m}; 0\mathrm{ m}$
0%
$800\mathrm{m},\ 800\sqrt{3}\mathrm{m}$
0%
$800\mathrm{m};400\sqrt{3}\mathrm{m}$
0%
$800; 800\sqrt{3}\mathrm{m}$

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:

0%
$\sqrt{72}$
0%
$\sqrt{33}$
0%
$\sqrt{45}$
0%
$\sqrt{18}$

Explanation

Take A as the origin(0,0)

$\vec {AB}= 3\hat i + 4\hat k$ 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

0%
$1:2$
0%
$3:2$
0%
$8:3$
0%
$4:3$

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)

$(\dfrac{2y_2+y_1}{2x_2+x_1})(x)-y_1=\dfrac{y_2-2y_1}{x_2-2x_1}(x-x_1)$

$x=\dfrac{(2x_2+x_1)}{5} =\dfrac{3(\dfrac{2x_2+x_1}{3})+2(0)}{(3+2)}$

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

0%
$\dfrac{1}{5}\vec{PR}$
0%
$\dfrac{3}{5}\vec{PR}$
0%
$\dfrac{2}{5}\vec{PR}$
0%
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

0%
$\displaystyle \sqrt { \dfrac { 136 }{ 9 } }$
0%
$\displaystyle \sqrt { \dfrac { 139 }{ 9 } }$
0%
$\displaystyle \dfrac { 20 }{ 3 }$
0%
$\displaystyle \sqrt { \dfrac { 217 }{ 9 } }$

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

0%
$3$
0%
$9$
0%
$7$
0%
$4$

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.

$\overrightarrow{b}$ 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

0%
$\left[-6, -\dfrac{1}{6}\right]$
0%
$\left(- \infty, -6 \right] \cup \left[-\dfrac{1}{6}, \infty \right)$
0%
$\left[0, 6 \right]$
0%
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

$25 (1 + k^2) \leq 37 (k^2 + 2k +1)$

$6k^2 + 37k + 6 \geq 0$ or

$(6k + 1) (k + 6) \geq 0$

$k \in (-\infty, - 6] \cup \left [ - \dfrac{1}{6}, \infty \right )$

$\vec{a},\vec{b},\vec{c}$ 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)

0%
$\vec{a}. \vec{b}=\vec{b}. \vec{c}=\vec{c} . \vec{a}=0$
0%
$\left | \vec{b} \right |=\left | \vec{c} \right |$
0%
$\vec{a}$ is a unit vector
0%
$\vec{c}$ 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

0%
$- \hat i - \hat k$
0%
$\hat i - 2 \hat j - \hat k$
0%
$2 \hat i + \hat j + \hat k$
0%
$\hat i + \hat k$

Explanation

$\vec{p}, \vec{q}, \vec{r}$ 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

0%
Not coplanar
0%
Coplanar but cannot form a triangle
0%
Coplanar and form a triangle
0%
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

0%
$\displaystyle \dfrac{1}{2}$
0%
$1$
0%
$\displaystyle \dfrac{3}{2}$
0%
$2$

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

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{2}$

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

0%
$\dfrac{1}{5}(2\vec {b}+\vec {c})$
0%
$\dfrac{1}{6}(2\vec {b}+\vec {c})$
0%
$\dfrac{1}{7}(\vec {b}+2\vec {c})$
0%
$\dfrac{1}{7}(2\vec {b}+\vec {c})$

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

$AQ$ 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 the position vector of point

$P$ is

0%
$\displaystyle \dfrac{|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$
0%
$\displaystyle \dfrac{3|\vec{a}| |\vec{c}|}{3 |\vec{c}| + 2 |\vec{a}|} \left ( \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{c}}{|\vec{c}|} \right )$
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

0%
$\dfrac{3}{4}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{1}{2}$

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

$\displaystyle \dfrac{\lambda \vec{c}}{\lambda + 1} = \dfrac{\mu \vec{d} + \vec{a}}{\mu + 1}$

$\displaystyle \left ( \frac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \frac{\mu}{\mu + 1} \right )\vec{d} + \left ( \frac{1}{\mu + 1} \right ) \vec{a}$

$\displaystyle \left (\dfrac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \dfrac{\mu}{\mu + 1} \right ) \left ( \dfrac{-\vec{b} + 2 \vec{c}}{3} \right ) + \left ( \dfrac{1}{\mu + 1} \right ) 2 \left ( \vec{b} - \vec{c} \right )$ ....(using

$(i)$ and

$(ii)$ )

$\displaystyle \left ( \frac{\lambda}{\lambda + 1} \right ) \vec{c}= \left ( \frac{6 - \mu}{3 (\mu + 1)} \right ) \vec{b} + \left ( \frac{2 \mu}{3 (\mu + 1)} - \frac{2}{\mu + 1} \right ) \vec{c}$

$\displaystyle \left ( \dfrac{\lambda}{\lambda + 1} \right ) \vec{c} = \left ( \dfrac{6 - \mu}{3 (\mu + 1)} \right ) \vec{b} + \left ( \dfrac{2\mu - 6}{3 (\mu + 1)} \right ) \vec{c}$

$\displaystyle \left ( \frac{6 - \mu}{3 (\mu + 1)} \right )\vec{b} + \left ( \frac{2 \mu - 6}{3 (\mu + 1)} - \frac{\lambda}{\lambda + 1} \right ) \vec{c} = \vec{0}$

$\displaystyle \dfrac{6 - \mu}{3 (\mu + 1)} = 0$ and

$\displaystyle \dfrac{2 \mu - 6}{3 (\mu + 1)} - \dfrac{\lambda}{\lambda + 1} = 0$

(as

$\vec{b}$ and

$\vec{c}$ are non-collinear)

$\displaystyle \mu = 6, \lambda = \dfrac{2}{5}$

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

0%
$\dfrac{4}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$

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

0%
$\displaystyle \bar{\beta }+\dfrac{\bar{\beta }-\bar{\alpha }}{\left ( \bar{\alpha } \right )^{2}}\bar{\alpha }$
0%
$\displaystyle \dfrac{\left ( \bar{\alpha }\times \bar{\beta } \right )\times \bar{\alpha }}{\left ( \bar{\alpha } \right )^{2}}$
0%
$\displaystyle \dfrac{\bar{\beta }\cdot \bar{\alpha }}{\left ( \alpha \right )^{2}}\bar{\alpha }+\bar{\beta }$
0%
$\displaystyle \left | \bar{\beta } \right |\dfrac{\bar{\alpha }\times \left ( \bar{\alpha }\times \bar{\beta } \right )}{\left ( \alpha \right )^{2}}$

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

0%
$2 \widehat{i} + 3 \widehat{j} - 3 \widehat{k}$
0%
$- 2 \widehat{i} - \widehat{j} + 5 \widehat{k}$
0%
$2 \widehat{i} + 3 \widehat{j} + 3 \widehat{k}$
0%
$2 \widehat{i} + \widehat{j} + 5 \widehat{k}$

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

$x_1 \in R$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 11 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 11 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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