Vector Algebra Test

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Vector Algebra Test - 56

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Question 1
1 / -0

If vectors $$\vec {A} = 2\hat {i} + 3\hat {j} + 4\hat {k}, \vec {B} = \hat {i} + \hat {j} + 5\hat {k}$$ and $$\vec {C}$$ form a left-handed system, then $$\vec {C}$$ is

A
$$11\hat {i} - 6\hat {j} - \hat {k}$$

B
$$-11\hat {i}+ 6\hat {j} + \hat {k}$$

C
$$11\hat {i} - 6\hat {j} + \hat {k}$$

D
$$-11\hat {i} + 6\hat {j} - \hat {k}$$

Solution
Question 2
1 / -0

Let $$\overrightarrow{a}=2\hat{i}-3\hat{j}+6\hat{k}$$ and $$\overrightarrow{b}=-2\hat{i}+3\hat{j}-\hat{k}$$ then projection of $$\overrightarrow{a}$$ on $$\overrightarrow{b} :$$ projection of $$\overrightarrow{b}$$ on $$\overrightarrow{a}=$$

A
$$3:7$$

B
$$7:3$$

C
$$-4$$

D
$$3$$

Solution

Projection of $$\overrightarrow{a}$$ on $$\overrightarrow{b}$$
$$ = $$ scalar components of $$\overrightarrow{a}$$ along $$\overrightarrow{b}$$
$$=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$$
Projection of $$\overrightarrow{b}$$ on $$\overrightarrow{a}$$
$$ = $$ scalar components of $$\overrightarrow{b}$$ along $$\overrightarrow{a}$$
$$ = \dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|}$$
Ratio $$= \dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}:\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|}$$
$$=\left|\overrightarrow{a}\right|:\left|\overrightarrow{b}\right|=7:3$$
where$$\left|\overrightarrow{a}\right|=\sqrt{4+9+36}=\sqrt{49}=7$$
$$\left|\overrightarrow{b}\right|=\sqrt{4+4+1}=\sqrt{9}=3$$
Thus, the ratio is $$7:3$$
Question 3
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If $$2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}=0 \Rightarrow \overrightarrow{a}\times \overrightarrow{b} + \overrightarrow{b}\times \overrightarrow{c} +\overrightarrow{c}\times \overrightarrow{a}=$$

A
$$0$$

B
$$3\overrightarrow{a}\times \overrightarrow{b}$$

C
$$3\overrightarrow{b}\times \overrightarrow{c}$$

D
$$3\overrightarrow{c}\times \overrightarrow{a}$$

Solution

$$2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}=0$$
$$\Rightarrow \left(2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}\right)\times \overrightarrow{a}=0$$
$$\Rightarrow 4\overrightarrow{c}\times \overrightarrow{a}=3\overrightarrow{a}\times \overrightarrow{b}$$ ..................$$\left(1\right)$$
$$\left(2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}\right)\times\overrightarrow{b}=0$$
$$\overrightarrow{a}\times \overrightarrow{b}=2\overrightarrow{b}\times \overrightarrow{c}$$ ............$$ \left(2\right)$$
$$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}$$
$$ = \overrightarrow{a}\times \overrightarrow{b}+\dfrac{1}{2}\overrightarrow{a}\times \overrightarrow{b}+\dfrac{3}{4}\overrightarrow{a}\times \overrightarrow{b}$$
$$ = \overrightarrow{a}\times \overrightarrow{b}\left(1+\dfrac{1}{2}+\dfrac{3}{4}\right)$$
$$ = \dfrac{9}{4}\overrightarrow{a}\times \overrightarrow{b}$$
$$ =3\times \dfrac{3}{4}\overrightarrow{a}\times \overrightarrow{b}$$
$$=3\overrightarrow{c}\times\overrightarrow{a}$$ using $$\left(1\right)$$ and $$ \left(2\right)$$
Question 4
1 / -0

If $$|\vec{a}|=5, |\vec{a}-\vec{b}|=8$$ and $$|\vec{a}+\vec{b}|=10$$, then $$|\vec{b}|$$ is equal to :

A
$$1$$

B
$$\sqrt{57}$$

C
$$13$$

D
$$14$$

Solution

$$\begin{array}{l} \left| { \vec { a } } \right| =5,\left| { \vec { a } -\vec { b } } \right| =8,\left| { \vec { a } +\vec { b } } \right| =10 \\ { \left| { \vec { a } -\vec { b } } \right| ^{ 2 } }=64 \\ { \left| { \vec { a } } \right| ^{ 2 } }+{ \left| { \vec { b } } \right| ^{ 2 } }-2\vec { a } \cdot \vec { b } =64 \\ { \left| { \vec { a } +\vec { b } } \right| ^{ 2 } }=100 \\ { \left| { \vec { a } } \right| ^{ 2 } }+{ \left| { \vec { b } } \right| ^{ 2 } }-2\vec { a } \cdot \vec { b } =100 \\ { \left| { \vec { b } } \right| ^{ 2 } }-2\vec { a } \cdot \vec { b } =39 \\ { \left| { \vec { b } } \right| ^{ 2 } }+2\vec { a } \cdot \vec { b } =75 \\ 2{ \left| { \vec { b } } \right| ^{ 2 } }=114 \\ { \left| { \vec { b } } \right| ^{ 2 } }=57 \\ \left| { \vec { b } } \right| =\sqrt { 57 } \\ Hence,\, correct\, option\, is\, \left( B \right) \end{array}$$
Question 5
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Let $$\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$$ . A vector in the plane of $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$, where projection on $$\overrightarrow{c}$$ is $$\dfrac{1}{\sqrt{3}}$$, is

A
$$4\hat{i}-\hat{j}+4\hat{k}$$

B
$$3\hat{i}+\hat{j}-3\hat{k}$$

C
$$2\hat{i}+\hat{j}$$

D
$$4\hat{i}+\hat{j}-4\hat{k}$$

Solution

A vector in the plane of $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is
$$\overrightarrow{a}+\lambda \overrightarrow{b} =\left(\hat{i}+2\hat{j}+\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)$$
$$ =\left(1+\lambda\right)\hat{i}+\left(2-\lambda\right)\hat{j}+\left(1+\lambda\right)\hat{k}$$
If projection on $$\overrightarrow{c}$$ is $$\dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \left(\overrightarrow{a}+\lambda \overrightarrow{b}\right).\dfrac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\pm \dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \left(\left(1+\lambda\right)\hat{i}+\left(2-\lambda\right)\hat{j}+\left(1+\lambda\right)\hat{k}\right)\dfrac {\left(\hat{i}+\hat{j}-\hat{k}\right)} {\sqrt{1+1+1}} =\pm \dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \dfrac{1+\lambda+2-\lambda-1-\lambda} {\sqrt{3}} =\pm \dfrac{1}{\sqrt{3}}$$
$$\Rightarrow 2-\lambda=\pm 1$$
$$\Rightarrow \lambda=2\pm 1$$$$\therefore \lambda=1$$ or $$3$$
If $$\lambda =1, \overrightarrow{a}+\lambda \overrightarrow{b}=\left(1+1\right)\hat{i}+\left(2-1\right)\hat{j}+\left(1+1\right)\hat{k}$$
$$ =2\hat{i}+\hat{j}+2\hat{k}$$
If $$\lambda=3, \overrightarrow{a}+\lambda \overrightarrow{b} = \left(1+3\right)\hat{i}+\left(2-3\right)\hat{j}+\left(1+3\right)\hat{k}$$
$$ =4\hat{i}-\hat{j}+4\hat{k}$$
Question 6
1 / -0

A person travels $$12\ km$$ in the southward direction and then travels $$5\ km$$ to the right and then travels $$15\ km$$ toward the right and finally travels $$5\ km$$ towards the east, how far is he from his starting place?

A
$$5.5\ kms$$

B
$$3\ km$$

C
$$13\ km$$

D
$$6.4\ km$$

Solution

REF.Image
$$\vec{OA}= 12km$$
$$\vec{AB}= 5km$$
$$\vec{BC}= 15 km$$
$$\vec{CD}= 5 km$$
To find $$\vec{OD}= ?$$
after moving from $$\bar{O}$$ to $$\bar{A},\bar{B},\bar{C},\bar{D}$$ (along these directions)
He reads north of stating point i.e. O.
$$\therefore \bar{OD}$$ is southwards (3 km) from D or
or $$\bar{OD}$$ is northwards (3 km) from O
Option B is correct
Question 7
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Let $$\overrightarrow{u}=\hat{i}+\hat{j},\overrightarrow{v}=\hat{i}-\hat{j},\overrightarrow{w}=\hat{i}+2\hat{j}+3\hat{k}$$ If $$\hat{n}$$ is a unit vector such that $$\overrightarrow{u}.\hat{n}=0$$ and $$\overrightarrow {v}.\hat{n}=0$$, then $$\left|\overrightarrow {w}.\hat{n}\right|=$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$\overrightarrow{u}.\hat{n}=0$$
$$\overrightarrow{v}.\hat{n}=0$$
$$\Rightarrow\hat{n}$$ is along $$\overrightarrow{u}\times\overrightarrow{v}$$
$$=\begin{bmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&0\\1&-1&0\end{bmatrix}=-2k$$
$$\therefore\hat{n}=\pm k$$ (unit vector)
$$\overrightarrow{w}.\hat{n}=\left(\hat{i}+2\hat{j}+3\hat{k}\right).\left(\pm\hat{k}\right)$$
$$=\pm 3$$
$$\Rightarrow\left|\overrightarrow{w}.\hat{n}\right|=3$$
Question 8
1 / -0

Let $$\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$$ . A vector in the plane of $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$, where projection on $$\overrightarrow{c}$$ is $$\dfrac{1}{\sqrt{3}}$$, is

A
$$4\hat{i}-\hat{j}+4\hat{k}$$

B
$$3\hat{i}+\hat{j}-3\hat{k}$$

C
$$2\hat{i}+\hat{j}$$

D
$$4\hat{i}+\hat{j}-4\hat{k}$$

Solution

A vector in the plane of $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is $$\overrightarrow{a}+\lambda \overrightarrow{b} =\left(\hat{i}+2\hat{j}+\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)$$
$$ =\left(1+\lambda\right)\hat{i}+\left(2-\lambda\right)\hat{j}+\left(1+\lambda\right)\hat{k}$$
If projection on $$\overrightarrow{c}$$ is $$\dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \left(\overrightarrow{a}+\lambda \overrightarrow{b}\right).\dfrac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\pm \dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \left(\left(1+\lambda\right)\hat{i}+\left(2-\lambda\right)\hat{j}+\left(1+\lambda\right)\hat{k}\right)\dfrac {\left(\hat{i}+\hat{j}-\hat{k}\right)} {\sqrt{1+1+1}} =\pm \dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \dfrac{1+\lambda+2-\lambda-1-\lambda} {\sqrt{3}} =\pm \dfrac{1}{\sqrt{3}}$$
$$\Rightarrow 2-\lambda=\pm 1$$
$$\Rightarrow \lambda=2\pm 1$$
$$\therefore \lambda=1$$,or $$3$$
If $$\lambda =1, \overrightarrow{a}+\lambda \overrightarrow{b}=\left(1+1\right)\hat{i}+\left(2-1\right)\hat{j}+\left(1+1\right)\hat{k}$$
$$ =2\hat{i}+\hat{j}+2\hat{k}$$
If $$\lambda=3, \overrightarrow{a}+\lambda \overrightarrow{b} = \left(1+3\right)\hat{i}+\left(2-3\right)\hat{j}+\left(1+3\right)\hat{k}$$
$$ =4\hat{i}-\hat{j}+4\hat{k}$$
Question 9
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let $$\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$$ be such that $$\left|\overrightarrow{u}\right|=1,\left|\overrightarrow{v}\right|=2,\left|\overrightarrow{w}\right|=3$$. If the projection of $$\overrightarrow{v}$$ along $$\overrightarrow{u}$$ is equal to the projection of $$\overrightarrow{w}$$ along $$\overrightarrow{u}$$ and $$\overrightarrow{v},\overrightarrow{w}$$ are perpendicular to each other, then$$\left|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right|=$$

A
$$2$$

B
$$\sqrt{17} $$

C
$$\sqrt{14}$$

D
$$\sqrt{15}$$

Solution

Projection of $$\overrightarrow{v}$$ along $$\overrightarrow{u} =\dfrac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|}=\overrightarrow{u}.\overrightarrow{v}$$
Projection of $$\overrightarrow{w}$$ along $$\overrightarrow{u} =\overrightarrow{w}.\overrightarrow{u}$$
$$\overrightarrow{v}\perp \overrightarrow{w} \Rightarrow \overrightarrow{v}.\overrightarrow{w}=0$$ and $$\overrightarrow{u}.\overrightarrow{v}=\overrightarrow{u}.\overrightarrow{w} \left(given\right)$$
Now, $${\left|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right|}^{2} ={\left|\overrightarrow{u}\right|}^{2}+{\left|\overrightarrow{v}\right|}^{2}+{\left|\overrightarrow{w}\right|}^{2}-2 \overrightarrow{u}.\overrightarrow{v}-2 \overrightarrow{v}.\overrightarrow{w}+2 \overrightarrow{u}.\overrightarrow{w}$$
$$ = {1}^{2}+{2}^{2}+{3}^{2} = 14$$ on simplification
$$\therefore \left|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right|=\sqrt{14}$$
Question 10
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$$D,E\ and \ F$$ are the mid-point of the sides $$BC,CA \ and \ AB$$ respectively of the triangle $$ABC$$. Which of the following is true?

A
$${ \xrightarrow { AB= } \xrightarrow { 2ED } }$$

B
$${ \xrightarrow { AB= } \xrightarrow { 2DE } }$$

C
$${ \xrightarrow { AB= } \xrightarrow { ED } }$$

D
$${ \xrightarrow { AB= } \xrightarrow { 2DF } }$$

Solution

$$ From\,the\,figure $$
$$ ED=DG\,\,\,............\left( i \right) $$
$$ EG=AB\,\,\,\,...........\left( ii \right) $$
$$ ED+DG=AB $$
$$ \Rightarrow ED+ED=AB\,\,\,\,\,\,\,\,\,\,\left( from\left( i \right) \right) $$
$$ 2ED=AB $$

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Vector Algebra Test - 56

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Vector Algebra Test - 56

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SHARING IS CARING

Question 1

$$11\hat {i} - 6\hat {j} - \hat {k}$$

$$-11\hat {i}+ 6\hat {j} + \hat {k}$$

$$11\hat {i} - 6\hat {j} + \hat {k}$$

$$-11\hat {i} + 6\hat {j} - \hat {k}$$

Solution

Question 2

$$3:7$$

$$7:3$$

$$-4$$

$$3$$

Solution

Question 3

$$0$$

$$3\overrightarrow{a}\times \overrightarrow{b}$$

$$3\overrightarrow{b}\times \overrightarrow{c}$$

$$3\overrightarrow{c}\times \overrightarrow{a}$$

Solution

Question 4

$$1$$

$$\sqrt{57}$$

$$13$$

$$14$$

Solution

Question 5

$$4\hat{i}-\hat{j}+4\hat{k}$$

$$3\hat{i}+\hat{j}-3\hat{k}$$

$$2\hat{i}+\hat{j}$$

$$4\hat{i}+\hat{j}-4\hat{k}$$

Solution

Question 6

$$5.5\ kms$$

$$3\ km$$

$$13\ km$$

$$6.4\ km$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 8

$$4\hat{i}-\hat{j}+4\hat{k}$$

$$3\hat{i}+\hat{j}-3\hat{k}$$

$$2\hat{i}+\hat{j}$$

$$4\hat{i}+\hat{j}-4\hat{k}$$

Solution

Question 9

$$2$$

$$\sqrt{17} $$

$$\sqrt{14}$$

$$\sqrt{15}$$

Solution

Question 10

$${ \xrightarrow { AB= } \xrightarrow { 2ED } }$$

$${ \xrightarrow { AB= } \xrightarrow { 2DE } }$$

$${ \xrightarrow { AB= } \xrightarrow { ED } }$$

$${ \xrightarrow { AB= } \xrightarrow { 2DF } }$$

Solution

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