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CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 1 - MCQExams.com
CBSE
Class 12 Commerce Maths
Vector Algebra
Quiz 1
If $$|\overrightarrow{C}|^2=60$$ and $$\overrightarrow{C} \times (\widehat{i}+2\widehat{j}+5\widehat{k})=\overrightarrow{0}$$, then a value of $$\overrightarrow{C}\cdot (-7 \widehat{i}+2\widehat{j}+3\widehat{k})$$ is :
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$$4\sqrt{2}$$
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$$24$$
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$$12\sqrt{2}$$
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$$12$$
Explanation
$$ \vec {C}$$ is parallel to $$\widehat{i}+2\widehat{j}+5\widehat{k}$$.
$$ \Rightarrow \vec {C}=t(\widehat{i}+2\widehat{j}+5\widehat{k})$$, where '$$t$$' is a scalar.
$$ |\vec{C}|^2=60$$
$$\Rightarrow t^2 ( \hat{i}.\hat{i} + 4 \hat{j}.\hat{j} + 25 \hat{k}.\hat{k} ) = 60 $$
$$ \Rightarrow 60 = t^2 \times 30 $$
$$\Rightarrow t= \pm \sqrt{2}$$
Now
$$ \vec{C}.(-7i+2\widehat{j}+3\widehat{k})= \pm \sqrt{2} (i + 2 \widehat{j}+5\widehat{k})\cdot (-7\widehat{i} +2\widehat{j}+3 \widehat{k} )$$
$$=\pm \sqrt{2}[-7+4+15]$$
$$= \pm 12 \sqrt{2}$$
In a parallelogram ABCD, $$|\overrightarrow{AB}| = a, |\overrightarrow{AD}| = b$$ and $$|\overrightarrow{AC}| = c$$, then $$\overrightarrow{DB}.\overrightarrow{AB}$$ has the value
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$$\displaystyle \frac{1}{2} (a^2 + b^2 + c^2)$$
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$$\displaystyle \frac{1}{4} (a^2 + b^2 - c^2)$$
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$$\displaystyle \frac{1}{3} (b^2 + c^2 - a^2)$$
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$$\displaystyle \frac{1}{2} (a^2 + b^2 - c^2)$$
Explanation
In a parallelogram ABCD, $$|\overrightarrow{AB}| = a, |\overrightarrow{AD}| = b$$ and $$|\overrightarrow{AC}| = c$$
$$c^2=a^2+b^2-2ab\cos B$$
$$\Rightarrow 2ab\cos B =-c^2+a^2+b^2$$
$$\vec{DB}\cdot \vec {AB}=(\vec {DA} + \vec {DC})\cdot\vec {AB}= ba \cos B + a^2=\displaystyle\frac{1}{2}(3a^2+b^2-c^2)$$
Hence, option D.
Let $$P,\ Q,\ R$$ and $$S$$ be the points on the plane with position vectors $$-2\hat{i}-\hat{j},\ 4\hat{i},\ 3\hat{i}+3\hat{j}$$ and $$-3\hat{i}+2\hat{j}$$ respectively. The quadrilateral $$PQRS$$ must be a
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parallelogram, which is neither a rhombus nor a rectangle
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square
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rectangle, but not a square
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rhombus, but not a square
Explanation
Evaluating midpoint of $$PR$$ and $$QS$$ which
gives $$M\equiv[\dfrac{\hat{i}}{2}+\hat{j}]$$, same for both.
$$
\vec{PQ}=\vec{SR}=6\hat{i}+\hat{j}
$$
$$
\vec{PS}=\vec{QR}=-\hat{i}+3\hat{j}
$$
$$
\Rightarrow \vec{PQ}\cdot\vec{PS}\neq 0
$$
$$\vec{PQ}\Vert\vec{SR},\ \vec{PS}\Vert\vec{QR}$$ and $$|\vec{PQ}|=|\vec{SR}|,\ |\vec{PS}|=|\vec{QR}|$$
Hence, $$PQRS$$ is a parallelogram but not rhombus or rectangle.
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Statement -$$1$$ is True, Statement -$$2$$ is True; Statement-$$2$$ is a correct explanation for Statement-$$1$$
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Statement -$$1$$ is True, Statement -$$2$$ is True; Statement-$$2$$ is NOT a correct explanation for Statement-$$1$$
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Statement -$$1$$ is True, Statement -$$2$$ is False
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Statement -$$1$$ is False, Statement -$$2$$ is True
Explanation
Since $$\overline{PQ}||\overline{TR}$$ .$$\cdot \overline{TR}$$ is resultant of $$\overline{SR}$$ and $$\overline{ST}$$ vector.
$$\Rightarrow\overline{PQ}\times(\overline{RS}+\overline{ST})\neq 0$$
But for statement $$2$$, we have $$\overline{PQ}\times\overline{RS}=\vec{0}$$
which is not possible as $$\overline{PQ}||\overline{RS}$$
Hence, statement $$1$$ is true and statement $$2$$ is false.
$$ABCD$$ is a parallelogram and $$AC, BD$$ be its diagonals Then $$ \vec{AC} +\vec{BD}$$ is
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$$2\vec{AB} $$
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$$2\vec{BC} $$
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$$3\vec{AB} $$
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$$3\vec{BC} $$
Explanation
$$ABCD$$ is a parallelogram so, $$AB||CD$$ and $$AD||BC$$
$$\vec{AC}=\vec{AB}+\vec{BC}$$ (addition Theorem)
$$\vec{BD}=\vec{BA}+\vec{AD}$$ (addition Theorem)
$$\vec{AC}+\vec{BD}=\vec{AB}+\vec{BC}+\vec{BA}+\vec{AD}$$
$$=\vec{AB}+\vec{BC}-\vec{AB}+\vec{BC}$$
$$=2\vec{BC}$$
The triangle $$ABC$$ is defined by the vertices $$A= (0,7,10)$$ , $$B=(-1,6,6)$$ and $$C=(-4,9,6)$$. Let $$D$$ be the foot of the attitude from $$B$$ to the side $$AC$$ then $$BD$$ is
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$$\overline{i}+2\overline{j}+2\overline{k}$$
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$$-\overline{i}+2\overline{j}+2\overline{k}$$
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$$\overline{i}+2\overline{j}-2\overline{k}$$
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$$\overline{i}-2\overline{j}+2\overline{k}$$
Explanation
Given : $$A=(0,7,10), B=(-1,6,6)$$ and $$C=(-4,9,6)$$
Position vector(P.V.) of $$AB$$ is $$(-1,-1,-4)$$
$$\implies |AB|=\sqrt{(-1)^2+(-1)^2+(-4)^2}=\sqrt{18}$$
P.V. of $$BC$$ is $$(-3,3,0)$$
$$\implies |BC|=\sqrt{(-3)^2+3^2+0}=\sqrt{18}$$
P.V. of $$AC$$ is $$(-4,2,-4)$$
$$\implies |AC|=\sqrt{(-4)^2+2^2+(-4)^2}=6$$
Now, $$D$$ is the midpoint of $$AC$$
$$\implies D=\left(\dfrac{0-4}{2}, \dfrac{7+9}{2}, \dfrac{10+6}{2}\right)= (-2,8,8)$$
$$\implies \overrightarrow{BD} = -\hat{i}+2\hat{j}+2\hat{k}$$
The point $$C=(\dfrac{12}{5}, \dfrac{-1}{5},\dfrac{4}{5})$$ divides the line segment $$AB$$ in the ratio $$3:2$$. If $$B=(2,-1,2)$$ then $$A$$ is
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$$
(3, 1, 1)$$
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$$(3, 1,-1)$$
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$$(3,-1,-1)$$
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$$(-3,1,-1)$$
Explanation
By S
ection Formula
$$C= \dfrac{3B+2A}{5}$$
$$(12,-1,4)= (6,-3,6)+2(A)$$
$$(6,2,-2)=2A$$
$$A=(3,1,-1)$$
Let $$\vec{A}= \hat{i}+6\mathrm{i}+6\mathrm{k},\vec{B}=-4\hat{i}+9\hat{i}+6\hat{k},\vec{G}=\displaystyle \dfrac{-5}{3}\hat{i}+\dfrac{22}{3}\hat{j}+\dfrac{22}{3}\hat{k}$$. If $$\mathrm{G}$$ is the centroid then the triangle $$ABC$$ is
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A right angled triangle
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A right angled isosceles triangle
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An isosceles triangle
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An equilateral triangle
Explanation
Centroid of a Triangle is given by $$G=\dfrac{A+B+C}{3}$$
Putting values
$$\dfrac{-5\hat{i}+15\hat{j}+12k+C}{3}= \dfrac{-5\hat{i}}{3}+\dfrac{22}{3}\hat{j}+\dfrac{22}{3}\hat{k}$$
$$\vec{C}= 7\hat{j}+10\hat{k}$$
$$BC^{2}= AC^{2}+AB^{2}$$ right angled at $$A$$
& $$AB=AC$$ (isosceles triangle)
If $$A(\overline{a})$$ , $$B(\overline{b})$$ and $$C(\overline{c})$$ be the vertices of a triangle $$ABC$$ whose circumcentre is the origin then orthocentre is given by
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$$\overline{ a}+\overline{b}+\overline{c}$$
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$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{3}$$
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$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{2}$$
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$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{4}$$
Explanation
P.V. of circumcentre $$= (0,0,0)$$
P.V. of circumcentre $$= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$$
$$centroid = \dfrac{orthocentre + 2\ circumcentre}{3}$$ (section formula)
$$\dfrac{\vec{a}+\vec{b}+\vec{c}}{3}= \dfrac{orthocentre}{3}$$
orthocentre$$ = \vec{a}+\vec{b}+\vec{c}$$
Let $$a,b,c,d $$ be the position vectors of the points $$\mathrm{A},\mathrm{B},\mathrm{C},\mathrm{D}$$ respectively. The condition for the figure $$ABCD$$ to be a parallelogram is
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$$\hat { a } +{ \hat { b } }={ \hat { c } }+\hat { d } $$
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$${ \hat { a } }+\hat { b } ={ \hat { c } }+{ \hat { d } }={ 0 }$$
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$${ \hat { a } }+{ \hat { c } }={ \hat { b } }+\hat { d } $$
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$${ \hat { a } }+{ \hat { c } }=\hat { b } +{ \hat { d } }={ 0 }$$
Explanation
For $$ABCD$$ to be a Parallelogram $$\vec{AB}||\vec{DC}$$ and $$\vec{AB}=\vec{DC}$$
So $$\vec{b}-\vec{a}=\vec{c}-\vec{d}$$
$$\vec{b}+\vec{d}=\vec{a}+\vec{c}$$ --- (1)
or $$\vec{AD}=\vec{BC}$$
$$\vec{d}-\vec{a}=\vec{c}-\vec{b}$$
$$\vec{d}+\vec{b}=\vec{c}+\vec{a}$$ --- (2)
Let $$2\hat{i}+\hat{k}=\vec{\mathrm{a}},\ 3\hat{j}+4\hat{k}=\vec{{b}}$$, $$8\hat{i}-3\hat{j}$$ $$=\vec{\mathrm{c}}$$. If $$\vec{a}={x}\vec{b}+{y}\vec{{c}}$$, then $$(x,y) $$ is equal to
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$$\left ( \dfrac{1}{2},\dfrac{1}{2} \right )$$
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$$\left ( \dfrac{1}{3},\dfrac{1}{3} \right )$$
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$$\left ( \dfrac{1}{4},\dfrac{1}{4} \right )$$
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$$(\displaystyle \dfrac{3}{4},\dfrac{3}{4})$$
Explanation
$$2\hat{i}+\hat{k}= x(3\hat{j}+4\hat{k})+y(8\hat{i}-3\hat{j})$$
$$8y= 2 $$ $$( as\ \hat{i},\ \hat{j},\ \hat{k}$$ are not collinear)
$$y=\dfrac{1}{4}$$
$$4x= 1$$
$$x=\dfrac{1}{4}$$
If $$G$$ is the centroid of the triangle $$ABC$$ then $$\vec{GA}+\vec{GB}+\vec{GC}$$ is equal to
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$$\vec{AB}$$
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$$\vec{BC}$$
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$$4\vec{GA}$$
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$$0$$
Explanation
Putting values in given expression
$$\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}= 3\vec{G}-(\vec{a}+\vec{b}+\vec{c})$$
$$= 3(\dfrac{\vec{a}+\vec{b}+\vec{c}}{3})-(\vec{a}+\vec{b}+\vec{c})$$
$$= 0$$
If the position vectors of the points $$A, B, C, D$$ are$$(0,2, 1)$$, $$(3,1,1),$$ $$(-5,3,2)$$,$$(2,4,1)$$ respectively and if $$PA+PB+PC+PD=0$$ then the position vector of P is
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$$(0,\displaystyle \dfrac{5}{2},\dfrac{5}{4})$$
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$$(\dfrac{5}{2},\dfrac{5}{2},\dfrac{5}{4})$$
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$$(\displaystyle \dfrac{5}{2},0,\dfrac{5}{4})$$
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$$(\displaystyle \dfrac{5}{2},\dfrac{5}{4},0)$$
Explanation
Taking $$P$$ common from given condition
$$4\vec{P}-(\vec{A}+\vec{B}+\vec{C}+\vec{D})= 0$$
$$\vec{P}= \dfrac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4}$$
$$\vec{P}=(0, \dfrac{10}{4}, \dfrac{5}{4})$$
Let $$\mathrm{A}\mathrm{B}\mathrm{C}$$ be a triangle and let $$\mathrm{S}$$ be its circumcentre and $$\mathrm{O}$$ be its orthocentre. The $$\overline{\mathrm{S}\mathrm{A}}+\overline{\mathrm{S}\mathrm{B}}+\overline{\mathrm{S}\mathrm{C}}= $$
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$$4\overline{\mathrm{S}\mathrm{O}}$$
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$$3\overline{\mathrm{S}\mathrm{O}}$$
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$$2\overline{\mathrm{S}\mathrm{O}}$$
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$$\overline{\mathrm{S}\mathrm{O}}$$
Explanation
$$\overline{SA}+\overline{SB}+\overline{SC}=(\bar{A}+\bar{B}+\bar{C})-3\bar{S}\ -(1)$$
$$S$$is circumcentere
$$C$$ is centroid
$$O$$ is orthocentre
$$2C+0=3S-3C$$
$$0-C = 3(S-C)$$
$$\overline{CO}=3\ \overline{CS}\ -(2)$$
apply (2) in (1)
$$=3(\frac{(\bar{A}+\bar{B}+\bar{C})}{3}-\bar{S})$$
$$=\ \overline{SO}$$
If $$\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$$, then
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$$\vec{\mathrm{a}}\neq\vec{b}$$
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$$\mathrm{\vec{a}}=k\mathrm{\vec{b}}$$
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This result is impossible
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This result is always true.
Explanation
Given,
$$\vec{a}\times \vec{b}=\vec{b}\times\vec{a}$$
$$\vec{a}\times \vec{b}=-(\vec{a}\times\vec{b})$$
$$\implies 2(\vec{a}\times\vec{b})=0$$
$$\implies \vec{a}\times \vec{b}=0$$
Hence, $$\vec{a}$$ is parallel to $$\vec{b}$$
$$\vec{a}=k\vec{b}$$, where $$k$$ is a scalar quantity.
Taking $$O$$' as origin and the position vectors of $$A, B$$ are $$\vec i+3\vec{j}-2\vec k, 3\vec{i}+\vec{j}-2\vec{k}$$. The vector $$\overrightarrow{OC}$$ is bisecting the angle $$AOB$$ and if $$C$$ is a point on line $$\overrightarrow{AB}$$ then $$C$$ is
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$$4(\vec{i}+\vec{j}-\vec{k})$$
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$$2(\vec{i}+\vec{j}-\vec{k})$$
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$$(\vec{i}+\vec{j}-\vec{k})$$
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$$6(\vec{i}+\vec{j}-\vec{k})$$
Explanation
Taking magnitude of vectors
$$\left | \overrightarrow{OA} \right |=\sqrt{(0-1)^2+(0-3)^2+(0+2)^2}= \sqrt{14}$$
and $$ \left | \overrightarrow{OB} \right |=\sqrt{(0-3)^2+(0-1)^2+(0+2)^2}= \sqrt{14}$$
So, $$\dfrac{AC}{CB}= \dfrac{1}{1}$$
Applying Section Formula
$$C= \dfrac{B+A}{2} $$
$$C=\dfrac{\vec i+3\vec{j}-2\vec k+3\vec{i}+\vec{j}-2\vec{k}}{2}$$
$$C= 2(\hat{i}+\hat{j}-\hat{k})$$
The resultant of two concurrent forces $$n\vec{OP}$$ and $$m\vec{OQ}$$ is $$(m+n)\vec{OR}$$. Then $$R$$ divides $$PQ$$ in the ratio
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$$m:n$$
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$$n:m$$
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$$1:n$$
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$$m:1$$
Explanation
Applying Section Formula
$$R= \dfrac{KQ+P}{K+1}$$
$$(K+1)R= KQ+P$$
$${K+1}= \dfrac{m+n}{n}$$
$${K}= \dfrac{m}{n}$$
$$P, Q, R, S$$ have position vectors $$\overline{p},\overline{q},\overline{r},\overline{s}$$ respectively such that $$\overline{p}-\overline{q}=2(\overline{s}-\overline{r})$$, then which of the following is correct
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$$PQ$$ and $$RS$$ bisect each other
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$$PQ$$ and $$PR$$ bisect each other
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$$PQ$$ and $$RS$$ trisect each other
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$$QS$$ and $$PR$$ trisect each other
Explanation
Given : $$P,Q,R,S$$ have position vectors $$\bar{p}, \bar{q}, \bar{r}, \bar{s}$$
$$\bar{p}-\bar{q}=2(\bar{s}-\bar{r})$$
$$\implies \bar{p}-\bar{q}=2\bar{s}-2\bar{r}$$
$$\implies \bar{p}+2\bar{r}=\bar{q}+2\bar{s}$$
$$\implies \dfrac{\vec{p}+2\vec{r}}{3}= \dfrac{\vec{q}+2\vec{s}}{3} $$
$$\implies \dfrac{\vec{P}+2\vec{R}}{3}= \dfrac{\vec{Q}+2\vec{S}}{3} $$
Thus, $$QS$$ & $$PR$$ trisect each other.
Let $$\vec{A}=\hat{i}+2\hat{j}{+}3\hat{k},\ \vec{B}=4\hat{i}+2\hat{j},\ \vec{C}=2\hat{i}+2\hat{j}{+}2\hat{k}$$. Then the ratio in which $$C$$ divides $$AB$$ is
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$$3:4$$
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$$1:3$$
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$$1:2$$
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$$1:1$$
Explanation
Let $$A=(1,2,3)$$, $$B=(4,2,0)$$ and $$C=(2,2,2)$$
Now let $$C$$ divide the line joining $$BA$$ in the ration $$M:N$$.
Then $$C_{Z}=\dfrac{MA_{Z}+NB_{Z}}{M+N}$$
$$\Rightarrow 2=\dfrac{M\times3+N\times0}{M+N}$$
$$\Rightarrow 2M+2N=3M$$
$$\Rightarrow 2N=M$$
$$\Rightarrow \dfrac{N}{M}=\dfrac{1}{2}$$
Hence, the required ratio is $$1:2$$.
If $$\overline{p}$$ is the position vector of the orthocentre and $$\overline{g}$$ is the position vector of the centroid of the triangle $$ABC$$ when circumcenter is the origin and if $$\overline{p}=\lambda\overline{g}$$ then $$\lambda=$$
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$$3$$
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$$2$$
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$$\displaystyle \dfrac{1}{3}$$
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$$\displaystyle \dfrac{2}{3}$$
Explanation
Let $$a,b,c$$ be position vectors of $$A,B,C$$ respectively.
$$\bar{p}$$ is the position vector of orthocentre and $$\bar{g}$$ is the position vector of the centroid.
$$\implies \bar{p}=\bar{a}+\bar{b}+\bar{c}$$ and $$\bar{g}=\dfrac{\bar{a}+\bar{b}+\bar{c}}{3}$$
$$\implies \vec{g} = \dfrac{\vec{p}}{3}\ \ (section\ formula)$$
$$\implies \bar{g}$$ trisect $$\bar{p}$$
$$\implies \vec{p} = 3\vec{g}$$
$$\implies \vec{p} = \lambda\vec{g}$$, where $$\lambda=3$$.
Hence, option A is correct.
Let $$A=(-3,4,-8)$$ and $$B=(5,-6,4)$$, then the coordinates of the point in which the $$XY$$- plane or $$XOY$$ plane divides the line segment $$AB$$ is
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$$(7,-8,0)$$
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$$(\displaystyle \dfrac{7}{3},-\dfrac{8}{3},0)$$
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$$(\displaystyle \dfrac{7}{2},-\dfrac{8}{2},0)$$
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$$(0,\displaystyle \dfrac{7}{3},-\dfrac{8}{3})$$
Explanation
Let $$\vec{O}$$ divide $$AB$$ into $$K:1$$
By Section Formula
$$(x,y,o)= \dfrac{KB+A}{K+1}$$
$$(x,y,o)= \dfrac{K(5,-6,4)+(-3,4,-8)}{K+1}$$
$$0= \dfrac{K(4)-8}{K+1}$$
$$K=2$$
$$x=\dfrac{10-3}{3}= \dfrac{7}{3}$$
$$y=\dfrac{-12+4}{3}= \dfrac{-8}{3}$$
If $$3\vec{a}+4\vec{b}-7\vec{c}=0$$ then the ratio in which $$C(\vec{c})$$ divides the join of $$A(\vec{a})$$ and $$B(\vec{b})$$ is
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$$1:2$$
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$$2:3$$
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$$3:2$$
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$$4:3$$
Explanation
Use section formula
$$C= \dfrac{K\vec{b}+\vec{a}}{K+1}$$
$$(K+1)\vec{c}= K\vec{b}+\vec{a}$$ ---- (1)
$$a+\dfrac{4}{3}\vec{b}-\dfrac{7}{3}\vec{c}= 0$$ -----(2)
$$\implies K= \dfrac{4}{3}$$
If $$\vec{a}+2\vec{b},2\vec{a}+\vec{b}$$ be the position vectors of the points $$A$$ and $$B$$, then the position vector of the point $$C$$ which divides $$AB$$ internally in the ratio $$2:1$$ is
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$$\displaystyle \dfrac{5\vec{a}-4\vec{b}}{3}$$
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$$\displaystyle \dfrac{5\vec{a}+4\vec{b}}{3}$$
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$$\displaystyle \dfrac{5\vec{a}-2\vec{b}}{3}$$
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$$\displaystyle \dfrac{5\vec{a}+2\vec{b}}{3}$$
Explanation
Applying Section Formula
$$C= \dfrac{2B+A}{3}$$
$$= \dfrac{4\vec{a}+2\vec{b}+\vec{a}+2\vec{b}}{3}$$
$$C= \dfrac{5\vec{a}+4\vec{b}}{3}$$
Hence, option B is correct.
In $$\Delta ABC$$, $$ P,\ Q,\ R$$ are points on $$BC,\ CA,\ AB$$ respectively, dividing them in the ratio $$1:4$$, $$3:2$$ and $$3 : 7$$. The point $$S$$ divides $$AB$$ in the ratio $$1:3$$. Then $$\displaystyle \dfrac{|\overline{AP}+\overline{BQ}+\overline{CR}|}{|\overline{CS}|}$$ is
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$$\displaystyle \dfrac{1}{5}$$
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$$\displaystyle \dfrac{2}{5}$$
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$$\displaystyle \dfrac{5}{2}$$
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$$\displaystyle \dfrac{7}{10}$$
Explanation
P.v of P $$= \dfrac{\vec{c}+4\vec{b}}{5}$$
P.v of R $$= \dfrac{7\vec{a}+3\vec{b}}{10}$$
P.v of Q$$= \dfrac{3\vec{a}+2\vec{b}}{5}$$
P.v of S $$= \dfrac{\vec{b}+3\vec{a}}{4}$$
$$\overrightarrow{AP}= \dfrac{\vec{c}+4\vec{b}-5\vec{a}}{5}\ \ \overrightarrow{BQ}= \dfrac{3\vec{a}+2\vec{c}-5\vec{b}}{5}$$
$$\overrightarrow{CR}= \dfrac{7\vec{a}+3\vec{b}-10\vec{a}}{10}\ \ \overrightarrow{CS}= \dfrac{\vec{b}+3\vec{a}-4\vec{c}}{4}$$
$$\therefore$$ $$\dfrac{\dfrac{\left | 2\vec{c}+4\vec{c}-10\vec{c}+8\vec{b}-10\vec{b}+3\vec{b}-10\vec{a}+6\vec{a}+7\vec{a} \right |}{10}}{\left | \dfrac{\vec{b}+3\vec{a}-4\vec{c}}{4} \right |}$$
$$= \dfrac{4}{10}$$
If $$\vec a=\hat i+2\hat j$$ and $$\vec b = 3\hat j$$, then $$\vec a\cdot\vec b=$$
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3
0%
-3
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6
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-6
Explanation
Yes, a triangular pyramid has 4 triangular faces.
If $$\overline{a}$$ and $$\overline{b}$$ are position vectors of $$A$$ and $$B$$ respectively, then the position vector of a point $$C$$ in $$AB$$ produced such that $$\overline{AC}=3\overline{AB}$$ is
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$$3\overline{a}-\overline{b}$$
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$$3\overline{b}-\overline{a}$$
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$$3\overline{a}-2\overline{b}$$
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$$3\overline{b}-2\overline{a}$$
Explanation
Let the Position Vector of $$C$$ be $$\vec{c}$$
We have $$\vec{AC}=3\vec{AB}$$
$$\vec{c} - \vec{a}= 3(\vec{b} - \vec{a})$$
$$\vec{c}= 3\vec{b} - 2\vec{a}$$
The projection of $$\displaystyle a=3i-j+5k$$ on $$\displaystyle b=2i+3j+k$$ is
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$$\displaystyle 8/\sqrt{\left ( 35 \right )}$$
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$$\displaystyle 8/\sqrt{\left ( 39 \right )}$$
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$$\displaystyle 8/\sqrt{\left ( 14 \right )}$$
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$$\displaystyle \sqrt{\left ( 14 \right )}$$
Explanation
Projection of $$a$$ on $$b$$ is $$=\cfrac{a\cdot b}{|b|} =\cfrac{3.2-1.3+5.1}{\sqrt{4+9+1}}=\cfrac{8}{ \sqrt{14}}$$
For $$O$$ being the origin and $$3$$ points $$P,Q$$ and $$R$$ lie on a plane. If $$\displaystyle \vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}$$, then $$P, Q, R$$ are
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the vertices of an equilateral triangle
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the vertices of an isoceles triangle
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collinear
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none of these
Explanation
$$\displaystyle \vec{PO}+\vec{OQ}=\vec{QO}+\vec{OR}$$
$$\Rightarrow \displaystyle -\vec{OP}+\vec{OQ}=-\vec{OQ}+\vec{OR}$$
$$\Rightarrow \displaystyle \vec{OQ}=\dfrac{\vec{OP}+\vec{OR}}{2}$$
Clearly $$Q$$ is the mid point of $$P$$ and $$R$$
Hence $$P,Q,R$$ are collinear.
If $$\left[ \overrightarrow { a } \overrightarrow { b } \overrightarrow { c } \right] =1$$ then $$\frac { \overrightarrow { a } .\overrightarrow { b }\times \overrightarrow { c } }{ \overrightarrow { c }\times \overrightarrow { a } .\overrightarrow { b } } +\frac { \overrightarrow { b } .\overrightarrow { c }\times \overrightarrow { a } }{ \overrightarrow { a }\times \overrightarrow { b } .\overrightarrow { c } } +\frac { \overrightarrow { c } .\overrightarrow { a }\times \overrightarrow { b } }{ \overrightarrow { b }\times \overrightarrow { c } .\overrightarrow { a } }$$
is equal to
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3
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1
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-1
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None of these
Explanation
$$\overrightarrow { a } .\overrightarrow { b } \times \overrightarrow { c } $$ is $$[\overrightarrow { a } \overrightarrow { b } \overrightarrow { c } ]$$
We snow that
$$[\overrightarrow { a } \overrightarrow { b } \overrightarrow { c } ]=[\overrightarrow { c } \overrightarrow { a } \overrightarrow { b } ]$$
Similarly,
$$[\overrightarrow { b } \overrightarrow { c } \overrightarrow { a } ]=[\overrightarrow { a } \overrightarrow { b } \overrightarrow { c } ]$$
= $$[\overrightarrow { c } \overrightarrow { a } \overrightarrow { b } ]$$
$$\therefore $$ Expression = 1 + 1 + 1
= 3
The projection $$\displaystyle 2\hat{i}+3\hat{j}-2\hat{k}$$ on the vector $$\displaystyle \hat{i}+2\hat{j}-3\hat{k}$$ is
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$$\displaystyle \frac{1}{\sqrt{14}}$$
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$$\displaystyle \sqrt{14}$$
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$$\displaystyle \frac{2}{\sqrt{14}}$$
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none of these
Explanation
Projection of $$\bar{a}$$ on $$\bar{b}$$ is a cos$${\theta}$$ where $$\theta$$ is the angle between the two.
Let $$\bar{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$$ and $$\bar{b} = \hat{i} + 2\hat{j} - 3\hat{k}$$
$$cos{\theta} = \dfrac{2 + 6 + 6}{\sqrt{17} \times \sqrt{14}}$$
Thus projection = $$(2 + 6 + 6)/\sqrt{14} = \sqrt{14}$$
Let $$\vec{AB}= 3\widehat {i} + \widehat{j}- \widehat{k}$$ and $$\vec{AC}=\widehat{i} -\widehat{j} +3\widehat{k}$$ and a point P on the line segment BC is equidistant from $$AB$$ and $$AC$$, then $$\vec{AP}$$ is
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$$2\widehat{i}- \widehat{k}$$
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$$\widehat{i}- 2\widehat{k}$$
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$$2\widehat{i} + \widehat{k}$$
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None of these
Explanation
Given : $$\vec{AB}=3\hat{i}+\hat{j}-\hat{k}$$ and $$\vec{AC}=\hat{i}-\hat{j}+3\hat{k}$$
$$P$$ is a point on $$BC$$ such that it is equidistant from $$AB$$ and $$AC$$
$$\implies P$$ is a midpoint of $$BC$$
$$\implies \vec{AP} = \cfrac{\vec{AB}+\vec{AC}}{2}$$
$$=\dfrac{4\hat{i}+2\hat{k}}{2}$$
$$ = 2\hat{i}+\hat{k}$$
Hence, option C is correct.
The vector sum of (N) coplanar forces, each of magnitude F, when each force is making an angle of
$$\frac{2\pi }{N}$$ with that preceding one, is :
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F
0%
$$\frac{NF}{2}$$
0%
NF
0%
Zero
Explanation
Given that:
Total forces$$ = N$$
An angle between forces$$=\dfrac{2\pi}{N}$$
Therefore,
Total angle between $$N$$ co-planer force $$=\dfrac{2\pi}{N}\times N = 2\pi$$
Thus, the forces represent a polygon of $$N$$ sides.
As, by the polygon method of vector addition, the vectors can be added by the placing the vectors in order as the sides of the polygon and the resultant of these vectors will join them to the tail of the first vector to
head of the last vector
. This will close the polygon and will give the magnitude and the direction of the resultant vector
.
As the $$N$$ number forces mentioned above, represents a closed polygon, the resultant will be zero.
If $$\vec{a}=3\hat{i}-2\hat{j}+\hat{k},\vec{b}=2\hat{i}-4\hat{j}-3\hat{k}$$, $$\vec{c}=-1\hat{i}+2\hat{j}+2\hat{k}$$ then $$\vec{a}+\vec{b}+\vec{c}=$$
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$$3\hat{i}-4\hat{j}$$
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$$3\hat{i}+4\hat{j}$$
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$$4\hat{i}-4\hat{j}$$
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$$4\hat{i}+4\hat{j}$$
Explanation
$$\vec{a}=3\hat{i}-2\hat{j}+\hat{k}$$
$$\vec{b}=2\hat{i}-4\hat{j}-3\hat{k}$$
$$\vec{c}=-1\hat{i}+2\hat{j}+2\hat{k}$$
$$\vec{a}+\vec{b}+\vec{c} = (3+2-1)\hat{i} + (-2-4+2)\hat{j} + (1-3+2)\hat{k} = 4\hat{i} - 4\hat{j}$$
Four point charges $${q}_A=2\mu C$$, $${q}_B=5\mu C$$, $${q}_C=2\mu C$$, $${q}_D=5\mu C$$ are located at the four corners of a square ABCD of side 10 cm. The force on a charge of $$1\mu C$$ placed at the centre of the square is
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Zero
0%
Towards AB
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Towards BC
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Towards AD
Explanation
$$q_B$$ will attract q along OB while $$q_D$$ towards OD having equal magnitude both will cancel each other's effect.
Similarly $$q_A$$ and $$q_C$$ will cancel each other's effect.
Thus net force can q will be zero
Namita walks 14 metres towards west, then turns to her right and walks 14 metres and then turns to her left and walks 10 metres. Again turning to her left she walks 14 metres. What is the shortest distance (in metres) between her starting point and the present position ?
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10
0%
24
0%
28
0%
38
Explanation
so, shortest distance $$=24$$
Hence,
option $$B$$ is correct answer.
If O is origin and C is the mid point of $$A(2, -1)$$ and $$B(-4,3)$$. Then value of OC is
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$$i + j$$
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$$-i + j$$
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$$i - j$$
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$$-i - j$$
Explanation
$$\text{Since C is mid point of A(2,-1) and B(-4,3).}$$
$$\therefore C(\dfrac{2-4}{2},\dfrac{-1+3}{2})i.e C(-1,1)$$
$$\therefore OC=(-1-0)\hat i+(1-0)\hat j=-\hat i+\hat j$$
Option B is correct.
A vector whose initial and terminal points coincide, is
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0%
Zero Vector
0%
Equal Vectors
0%
Null Vector
0%
Unit Vector
Explanation
The vector whose initial and terminals points are coincide has the length $$0$$. Hence we call it to be a zero vector and the zero vector no has the particular direction, so that it can be assigned in any direction.
If $$\vec{a}, \vec{b}, \vec{c}$$ are three non coplanar vectors, then $$(\vec{a}+\vec{b}+\vec{c})[(\vec{a}+\vec{b}) \times (\vec{a}+\vec{c})] $$ is :
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0
0%
$$2 [\vec{a}. \vec{b}. \vec{c}]$$
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$$- [\vec{a}. \vec{b}. \vec{c}]$$
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$$ [\vec{a}. \vec{b}. \vec{c}]$$
Explanation
Since$$\vec{a}, \vec{b}, \vec{c}$$ are three non-coplanar vectors. Then say
$$(\vec{a}+\vec{b}+\vec{c}) [(\vec{a}+\vec{b}) \times (\vec{a}+\vec{c})]=\vec{x}$$
$$\Rightarrow \vec{x}=(\vec{a}+\vec{b}+\vec{c}) [(\vec{a}\times \vec{a}+ \vec{a} \times \vec{c}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c}]$$
$$=(vec{a}+\vec{b}+\vec{c})[\vec{a}\times \vec{a}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c}]$$
$$[\because \vec{a}\times \vec{a}=0]$$
As $$[\vec{a} \vec{b} \vec{c}] =\vec{a}. (\vec{b} \times \vec{c})$$, so we can write
$$\vec{x}=[\vec{a}\vec{a}\vec{c}] + [\vec[{a}\vec{b} \vec{a}]+ [\vec{a} \vec{b}\vec{c}] + [\vec{b}\vec{a}\vec{c}] + [\vec{b} \vec{b} \vec{a}]+ [\vec{c} \vec{a} \vec{c}]+[\vec{c}\vec{b} \vec{c}]$$
Using $$[\vec{c}\vec[a]\vec{a}] =[\vec{b}\vec{c}\vec{a}]$$
$$=[\vec{c}\vec{a}\vec{b}]$$ & $$[\vec{a} \vec{a} \vec{c}] = [\vec{a} \vec{b} \vec{a}]$$
$$[\vec{c}\vec{a} \vec{a}]=0$$
$$\vec{x} = -[\vec{a}\vec{b} \vec{c}]$$
ABCDEF is a regular hexagon . Let $$\displaystyle \vec { AB } =a$$ and $$\displaystyle \vec { BC } =b$$. Express the vectors $$\displaystyle \vec { AC } $$ in terms if a and b.
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$$\vec{a}+\vec{b}$$
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$$\vec{a}-\vec{b}$$
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$$2\vec{a}$$
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None of these
Explanation
By the triangle law addition,
In triangle $$ABC$$
by this law sum of two vector are equal to third vector which is connecting the the tail of vector $$a$$ and head of $$b$$.
So therefore,
$$\vec {AC}=\vec a+\vec b$$
Given that u is a vector of length $$2$$, v is a vector of length $$3$$ and the angle between them when placed tail to tail is $$\displaystyle 45^{\circ} $$, which option is closest to the exact value of $$\vec u\cdot\vec v$$ ?
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$$4.5$$
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$$6.2$$
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$$4.2$$
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$$5.1$$
Explanation
$$u.v=|u||v|\cos (45^0)$$
$$=2.3.\cos(45^0)$$
$$=4.24$$
Option C is correct.
Six vectors, a through f have the magnitudes and directions indicated in the figure. Which of the following statements is true?
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$$b+e=f$$
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$$b+c=f$$
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$$d+c=f$$
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$$d+e=f$$
Explanation
Consider the problem
The $$f$$ vector has the resultant of any two other vector.
By parallelogram law of vector addition.
Let $$f$$ be the resultant between $$d$$ and $$e$$
Therefore,
$$d+e=f$$
Given the points $$A(-2,3,4),B(3,2,5),C(1,-1,2),\,\&\,D(3,2,-4)$$ The projection of the vector $$\displaystyle \underset{AB}{\rightarrow}$$ on the vector $$\displaystyle \underset{CD}{\rightarrow}$$ is
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$$\displaystyle \frac{22}{3}$$
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$$\displaystyle -\frac{21}{4}$$
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$$\displaystyle \frac{1}{7}$$
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$$-47$$
Explanation
The projection of the vector $$\displaystyle \underset{AB}{\rightarrow}$$ on the vector $$\displaystyle \underset{CD}{\rightarrow}$$ is $$\displaystyle \frac { \vec { AB } .\vec { CD } }{ \left| \vec { CD } \right| } =\frac { \left( 5i-j+k \right) .\left( 2i+3j-6k \right) }{ \left| \left( 2i+3j-6k \right) \right| } =\frac { 1 }{ 7 } $$
A zero vector has
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Any direction
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Many directions
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No direction
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None of these
Explanation
A zero or null vector have any or no direction.
Option A are C are correct.
If $$2\vec{a}+3\vec{b}-5\vec{c}=\vec{0}$$, then ratio in which $$\vec{c}$$ divides $$\vec{AB}$$ is
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$$3:2$$ internally
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$$3:2$$ externally
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$$2:3$$ internally
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$$2:3$$ externally
Explanation
Given, $$2\vec{a}+3\vec{b}-5\vec{c}=0$$
$$\Rightarrow \displaystyle\frac{2\vec{a}+3\vec{b}}{5}=\vec{c}$$
$$\Rightarrow \displaystyle\frac{2\vec{a}+3\vec{b}}{2+3}=\vec{c}$$
$$\Rightarrow \displaystyle\frac{\vec{a}+\displaystyle\frac{3}{2}\vec{b}}{1+\displaystyle\frac{3}{2}}=\vec{c}$$ .........$$(i)$$
Let $$\vec{c}$$ divide $$\vec{AB}$$ in the ratio $$\lambda :1$$
Then, $$\vec{c}=\displaystyle\frac{\vec{a}+\lambda \vec{b}}{1+\lambda}$$ ....$$(ii)$$
On comparing Eqs. $$(i)$$ and $$(ii)$$, we get $$\lambda=\displaystyle\frac{3}{2}$$
$$\therefore$$ Required ratio is $$3:2$$ internally.
Let $$ABCD$$ be a parallelogram whose diagonals intersect at $$P$$ and $$O$$ be the origin, then $$\vec { OA } +\vec { OB } +\vec { OC } +\vec { OD } $$ equals
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0%
$$\vec { OP } $$
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$$2\vec { OP } $$
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$$3\vec { OP } $$
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$$4\vec { OP } $$
Explanation
Consider the problem
Since
$$P$$ which is the intersection of diagonals of parallelogram its bisects the diagonal
Thus
$$OP=\frac{(OA+OC)}{2}$$
i.e.
$$OA+OC=2\,OP$$ ----- (i)
Similarly
$$OB+OD=OP$$ ----- (ii)
Adding (i) and (ii)
we get
$$\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD} = 4\overrightarrow {OP} $$
Hence the option $$D$$ is the correct answer.
The system of vectors $$i, j, k$$ is
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Orthogonal
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Collinear
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Coplanar
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None of these
Explanation
$$\text{Since i,j, k represent unit vector in the direction of X,Y and Z axis respectively.}$$
$$\text{Therefore, they are orthogonal.}$$
Option A is correct.
The projection of $$\displaystyle \overset { \rightarrow }{ a } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } -2\overset { \wedge }{ k } $$ on $$\displaystyle \overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } $$ is:
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$$\displaystyle \frac { 1 }{ \sqrt { 14 } } $$
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$$\displaystyle \frac { 2 }{ \sqrt { 14 } } $$
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$$\displaystyle \frac{-1}{ \sqrt { 14 } }$$
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$$\displaystyle \frac { -2 }{ \sqrt { 14 } } $$
Explanation
Projection of $$\displaystyle \overset { \rightarrow }{ a } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } -2\overset { \wedge }{ k } $$ on $$\displaystyle \overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } $$ is
$$=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\dfrac{2}{\sqrt {14}}$$
If $$\vec {a}$$ is a nonzero vector of magnitude $$'a'$$ and $$\lambda$$ a nonzero scalar, then $$\lambda{\vec {a}}$$ is unit vector if
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$$\lambda=1$$
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$$\lambda=-1$$
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$$a=|\lambda|$$
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$$a=\dfrac {1}{|\lambda|}$$
Explanation
Vector $$\lambda \vec {a}$$ is a unit vector if $$|\lambda \vec {a}|=1$$.
Now,
$$|\lambda \vec {a}|=1$$
$$\Rightarrow |\lambda||\vec {a}|=1$$
$$\Rightarrow |\vec {a}|=\dfrac {1}{|\lambda|}, \; [\lambda \neq 0]$$
$$\Rightarrow a=\dfrac {1}{|\lambda|}, \; [\because |\vec {a}|=a]$$
Hence, vector $$\lambda \vec {a}$$ is a unit vector if $$a=\dfrac {1}{|\lambda|}$$
Thus the correct answer is $$D$$.
Find $$u+v$$, when $$u=(3,4,-2)$$ and $$v=(0,-4,0)$$.
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$$(3,-8,-2)$$
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$$(-3,0,2)$$
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$$(3,0,-2)$$
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None of these
Explanation
Given points
$$u(3,4,-2)$$ and $$v(0,-4,0)$$
$$\vec{u}=3\hat{i}+4\hat{j}-2\hat{k}$$
$$\vec{v}=-4\hat{j}$$
vector
$$\vec{u}+\vec{v}=3\hat{i}+4\hat{j}-2\hat{k}-4\hat{j}$$
$$\vec{u}+\vec{v}=3\hat{i}-2\hat{k}=(3,0,-2)$$
Find the vector $$w$$ with the initial point $$(9,4)$$ and final point $$(12,6)$$.
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$$(21,10)$$
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$$(3,2)$$
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$$(-21,2)$$
0%
none of these
Explanation
Given end points
$$A(9,4)$$ and $$B(12,6)$$
$$\vec{OA}=9\hat{i}+4\hat{j}$$
$$\vec{OB}=12\hat{i}+6\hat{j}$$
vector w from A to B
$$\vec{AB}=\vec{OB}-\vec{OA}$$
$$\vec{w}=12\hat{i}+6\hat{j}-(9\hat{i}+4\hat{j})$$
$$\vec{w}=12\hat{i}+6\hat{j}-9\hat{i}-4\hat{j}$$
$$\vec{w}=3\hat{i}+2\hat{j}$$
$$=(3,2)$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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