Vector Algebra Test

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Vector Algebra Test - 13

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Question 1
1 / -0

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

A
$$\hat { a } +{ \hat { b } }={ \hat { c } }+\hat { d } $$

B
$${ \hat { a } }+\hat { b } ={ \hat { c } }+{ \hat { d } }={ 0 }$$

C
$${ \hat { a } }+{ \hat { c } }={ \hat { b } }+\hat { d } $$

D
$${ \hat { a } }+{ \hat { c } }=\hat { b } +{ \hat { d } }={ 0 }$$

Solution

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)
Question 2
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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

A
$$\left ( \dfrac{1}{2},\dfrac{1}{2} \right )$$

B
$$\left ( \dfrac{1}{3},\dfrac{1}{3} \right )$$

C
$$\left ( \dfrac{1}{4},\dfrac{1}{4} \right )$$

D
$$(\displaystyle \dfrac{3}{4},\dfrac{3}{4})$$

Solution

$$2\hat{i}+\hat{k}= x(3\hat{j}+4\hat{k})+y(8\hat{i}-3\hat{j})$$
$$8y= 2 $$ $$( as\ \hat{i},\ \hat{j},\ \hat{k}$$ are not collinear)
$$y=\dfrac{1}{4}$$
$$4x= 1$$
$$x=\dfrac{1}{4}$$
Question 3
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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

A
$$\overline{ a}+\overline{b}+\overline{c}$$

B
$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{3}$$

C
$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{2}$$

D
$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{4}$$

Solution

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$$ = \vec{a}+\vec{b}+\vec{c}$$
Question 4
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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

A
A right angled triangle

B
A right angled isosceles triangle

C
An isosceles triangle

D
An equilateral triangle

Solution

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)
Question 5
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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

A
$$(0,\displaystyle \dfrac{5}{2},\dfrac{5}{4})$$

B
$$(\dfrac{5}{2},\dfrac{5}{2},\dfrac{5}{4})$$

C
$$(\displaystyle \dfrac{5}{2},0,\dfrac{5}{4})$$

D
$$(\displaystyle \dfrac{5}{2},\dfrac{5}{4},0)$$

Solution

Taking $$P$$ common from given condition
$$4\vec{P}-(\vec{A}+\vec{B}+\vec{C}+\vec{D})= 0$$
$$\vec{P}= \dfrac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4}$$

$$\vec{P}=(0, \dfrac{10}{4}, \dfrac{5}{4})$$
Question 6
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If $$G$$ is the centroid of the triangle $$ABC$$ then $$\vec{GA}+\vec{GB}+\vec{GC}$$ is equal to

A
$$\vec{AB}$$

B
$$\vec{BC}$$

C
$$4\vec{GA}$$

D
$$0$$

Solution

Putting values in given expression
$$\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}= 3\vec{G}-(\vec{a}+\vec{b}+\vec{c})$$
$$= 3(\dfrac{\vec{a}+\vec{b}+\vec{c}}{3})-(\vec{a}+\vec{b}+\vec{c})$$
$$= 0$$
Question 7
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Let $$\mathrm{A}\mathrm{B}\mathrm{C}$$ be a triangle and let $$\mathrm{S}$$ be its circumcentre and $$\mathrm{O}$$ be its orthocentre. The $$\overline{\mathrm{S}\mathrm{A}}+\overline{\mathrm{S}\mathrm{B}}+\overline{\mathrm{S}\mathrm{C}}= $$

A
$$4\overline{\mathrm{S}\mathrm{O}}$$

B
$$3\overline{\mathrm{S}\mathrm{O}}$$

C
$$2\overline{\mathrm{S}\mathrm{O}}$$

D
$$\overline{\mathrm{S}\mathrm{O}}$$

Solution

$$\overline{SA}+\overline{SB}+\overline{SC}=(\bar{A}+\bar{B}+\bar{C})-3\bar{S}\ -(1)$$
$$S$$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)
$$=3(\frac{(\bar{A}+\bar{B}+\bar{C})}{3}-\bar{S})$$
$$=\ \overline{SO}$$
Question 8
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If $$\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$$, then

A
$$\vec{\mathrm{a}}\neq\vec{b}$$

B
$$\mathrm{\vec{a}}=k\mathrm{\vec{b}}$$

C
This result is impossible

D
This result is always true.

Solution

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.
Question 9
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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

A
$$4(\vec{i}+\vec{j}-\vec{k})$$

B
$$2(\vec{i}+\vec{j}-\vec{k})$$

C
$$(\vec{i}+\vec{j}-\vec{k})$$

D
$$6(\vec{i}+\vec{j}-\vec{k})$$

Solution

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

$$C= \dfrac{B+A}{2} $$

$$C=\dfrac{\vec i+3\vec{j}-2\vec k+3\vec{i}+\vec{j}-2\vec{k}}{2}$$

$$C= 2(\hat{i}+\hat{j}-\hat{k})$$
Question 10
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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 $$\lambda=$$

A
$$3$$

B
$$2$$

C
$$\displaystyle \dfrac{1}{3}$$

D
$$\displaystyle \dfrac{2}{3}$$

Solution

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.