Vector Algebra Test

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Vector Algebra Test - 74

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Question 1
1 / -0

If $$\hat{u}$$ and $$\hat{v}$$ are unit vectors and $$\theta$$ is the acute angle between them , then 2$$\hat{u}$$ 3$$\hat{v}$$ is a unit vector for

A
No value of $$\theta$$

B
Exactly one value of $$\theta$$

C
Exactly two values of $$\theta$$

D
More than two values of $$\theta$$

Solution

Given,
$$\bar{u}$$ and $$\vec{v}$$ are unit vector
$$|\vec{u}|=1$$
$$|\vec{v}|=1$$

Now,
$$\Rightarrow|2 \vec{u}||\overrightarrow{3 v}| \sin \theta=1$$
$$|\overrightarrow{2 u} \times \rightarrow{\vec{3 v}}|$$
$$[\because|\vec{a} \times \vec{b}|=|\vec a|\vec b|| \sin \theta]$$

$$\Rightarrow 2 \cdot 3|\bar{u}| \nabla \mid \sin \theta=1 \\$$

$$\sin \theta=\frac{1}{6}$$

$$\text { There is only one value of } \theta \\$$
$$\text { for which }|\vec 24 \times \vec 3V| \text { is a } \\$$
$$\text { unit vector. }$$

$$\text { option } B$$
Question 2
1 / -0

Let $$\vec{a}$$ = $$\hat{i}$$ $$\hat{j}$$, $$\vec{b}$$ = $$\hat{j}$$ $$\hat{k}$$, $$\vec{c}$$ = $$\hat{k}$$ $$\hat{i}$$. If $$\vec{d}$$ is a unit vector such that $$\vec{a}.\vec{d}$$ = 0 = $$\left | \vec{b}\vec{c}\vec{d} \right |$$ then $$\vec{d}$$ equals :

A
$$\dfrac{\hat{i} + \hat{j} 2\hat{k}}{\sqrt{6}}$$

B
$$\dfrac{\hat{i} + \hat{j} \hat{k}}{\sqrt{3}}$$

C
$$\dfrac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$$

D
$$ \hat{k}$$
Question 3
1 / -0

Vector $$\vec{ A} $$ and $$\vec{ B} $$ include an angle $$\theta $$ between them. If $$\vec { A}+\vec { B}$$ and $$\vec { A}-\vec { B}$$ respectively subtend angles $$\alpha $$ and $$\beta$$ with $$\vec { A}$$, then $$(\tan { \theta } +\tan { \beta} ) $$ is

A
$$\dfrac { (AB\sin { \theta})}{(A^2+B^2cos^2 \theta ) } $$

B
$$\dfrac { (2AB\sin { \theta})}{(A^2-B^2cos^2 \theta ) } $$

C
$$\dfrac { (A^2 \sin { \theta})}{(A^2+B^2cos^2 \theta ) } $$

D
$$\dfrac { (B^2\sin { \theta})}{(A^2+B^2cos^2 \theta ) } $$

Solution
Question 4
1 / -0

If $$a$$,$$b$$ and care three mutually perpendicular vectors, then the projection of the vector $$ \left|\frac{a}{|a|}+m \frac{b}{|b|}+n \frac{(a \times b)}{|a \times b|}\right. $$ along the angle bisector of the vector $$a$$ and $$b$$ is

A
$$ \frac{l^{2}+m^{2}}{\sqrt{l^{2}+m^{2}-n^{2}}} $$

B
$$ \sqrt{1^{2}+m^{2}+n^{2}} $$

C
$$ \frac{\sqrt{1^{2}+m^{2}}}{\sqrt{l^{2}-m^{2}+n^{2}}} $$

D
$$ \frac{1+m}{\sqrt{2}} $$

Solution
Question 5
1 / -0

The value of $$\lambda $$ for $$\left( x,y,z \right) \neq \left( 0,0,0 \right) $$ and $$\left( i+j+3k \right) x+\left( 3i-3j+k \right) y+\left( -4i+5j \right) z=\lambda \left( xi+yj+zk \right) $$ are

A
0, -1

B
0, 1

C
-2, 0

D
0, 2

Solution
Question 6
1 / -0

The position vector of a point lying on the joining the points whose position vectors are $$\overline i + \overline j -\overline k$$ and $$\overline i - \overline j +\overline k$$ is

A
$$\overline j$$

B
$$\overline i$$

C
$$\overline k$$

D
$$\overline 0$$

Solution
Question 7
1 / -0

If $$\overline { a } =-2\overline { i } +3\overline { j } +4\overline { k } $$ and $$\overline { b } =-2\overline { i } -2\overline { j } +3\overline { k } $$ then $$\overline { a } .\overline { b } $$ is

A
2

B
-2

C
6

D
none of these

Solution
Question 8
1 / -0

If $$\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } $$ are unit vectors such that $$\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0$$ then the value of $$\overset { \rightarrow }{ a. } \overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a. } $$ is

A
1

B
-1

C
-3/2

D
none of these

Solution
Question 9
1 / -0

Let position vector of the orthocentre of $$\triangle ABC$$ be $$\overrightarrow{r}$$. then, which of the following statement(s) is\are correct (Given position vector of points $$a\hat{i},b\hat{j},c\hat{k}$$ and $$abc=0$$)

A
$$\displaystyle \overrightarrow { r } .\bar { i } =\frac{a}{\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}}$$

B
$$\displaystyle \overrightarrow { r } .\bar { i } =\frac{1}{a\left(\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}} \right) }$$

C
$$\displaystyle \frac {\overrightarrow { r } .\bar { i }}{\overrightarrow { r } .\bar { j }}+\frac {\overrightarrow { r } .\bar { j }}{\overrightarrow { r } .\bar { k }}+\frac {\overrightarrow { r } .\bar { k }}{\overrightarrow { r } .\bar { i }}=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$

D
$$\displaystyle \frac {\overrightarrow { r } .\bar { i }}{\overrightarrow { r } .\bar { j }}+\frac {\overrightarrow { r } .\bar { j }}{\overrightarrow { r } .\bar { k }}+\frac {\overrightarrow { r } .\bar { k }}{\overrightarrow { r } .\bar { i }}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$

Solution
Question 10
1 / -0

If C is the mid point of AB and P is any point outside AB , then

A
$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ + $$\overrightarrow{PC}$$ = 0

B
$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ + $$2\overrightarrow{PC}$$ = $$\overrightarrow{0}$$

C
$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ = $$\overrightarrow{PC}$$

D
$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ = $$2\overrightarrow{PC}$$

Solution