Vector Algebra Test

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Vector Algebra Test - 51

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Question 1
1 / -0

Given: $$\vec a$$ and $$\vec b$$ are unit vector, and $$\theta $$ be the angle between them.
Then $$\dfrac{{1 - \vec a.\vec b}}{{1 + \vec a.\vec b}} = $$

A
$${\sin ^2}\frac{\theta }{2}$$

B
$${\cos ^2}\frac{\theta }{2}$$

C
$${\tan ^2}\frac{\theta }{2}$$

D
$${\cot ^2}\frac{\theta }{2}$$

Solution

$$\vec a \cdot \vec b = \cos \theta $$
$$\vec a \cdot \vec b = \cos \theta $$
$$\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }} = {\tan ^2}\frac{\theta }{2}$$
Question 2
1 / -0

Given that P = 12, Q = 5 and R = 13 also $$\vec P + \vec Q = \vec R$$, then the angle between $$\vec P$$ and $$\vec Q$$ will be :

A
$$

{\pi}

$$

B
$$

{\pi}\over{2}

$$

C
zero

D
$$

{\pi}\over{4}

$$

Solution

Consider the problem
Let $$x$$ is the angle between $$P$$ and $$Q$$ .

$$\vec R=\vec P+\vec Q $$

$$\therefore 13=\sqrt { ({ 5^{ 2 } }+{ { 12 }^{ 2 } }+(2{ { \times } }5{ { \times } }12\times Cosx)) } $$ [$$\because|\vec a+\vec b|=\sqrt{a^2+b^2+2ab\cos\theta}$$ ]

$$\Rightarrow 169=169{ { + } }120\cos x $$

$$\Rightarrow 0=120\cos x $$

$$\Rightarrow \cos x=0 $$

$$\therefore x=90^{\circ}=\dfrac{\pi}{2}$$

Hence option $$B$$ is the correct answer.
Question 3
1 / -0

A vector of magnitude 5 and perpendicular to $$\hat i - 2\hat j + \hat k$$ and $$2\hat i + \hat j - 3\hat k$$ is

A
$${{5\sqrt 3 } \over 3}\left( {\hat i + \hat j - \hat k} \right)$$

B
$${{5\sqrt 3 } \over 3}\left( {\hat i + \hat j + \hat k} \right)$$

C
$${{5\sqrt 3 } \over 3}\left( {\hat i - \hat j + \hat k} \right)$$

D
$${{5\sqrt 3 } \over 3}\left( {-\hat i + \hat j + \hat k} \right)$$

Solution

Let the vectors
$$\begin{array}{l}\vec a=\hat { i } -2\hat { j } +\hat { k } \\\vec b=2\hat { i } +\hat { j } -3\hat { k } \end{array}$$
and the vector perpendicular to $$a$$ and $$b$$ let it is $$c$$.
Now, for perpendicular
$$\vec c=\vec a \times \vec b$$
$$\vec { a } \times \vec { b } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i } } & { \hat { j } } & { \hat { k } } \\ 1 & { -2 } & 1 \\ 2 & 1 & { -3 } \end{array} } \right| $$
$$ = 5\hat i + 5\hat j + 5\hat k$$
$$\left| {\vec a \times \vec b} \right| = \sqrt {25 + 25 + 25} = 5\sqrt 3 $$
Unit vector along $$c$$ is $$\frac{1}{5\sqrt 3}(5 \hat i+5 \hat j+5\hat k)=\frac{1}{\sqrt3}(\hat i+\hat j+\hat k)$$
and and vector of magnitude $$5$$
$$=\frac{5}{\sqrt 3}(\hat i+\hat j+\hat k) = \frac{{5\sqrt 3 }}{3}\left( {\hat i + \hat j + \hat k} \right)$$
Question 4
1 / -0

The vertices of a triangle are $$A(1,1,2),B(4,3,1)$$ and $$C(2,3,5).$$ A vector representing the internal bisector of the angle $$A$$ id

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

B
$$2\hat{i}-2\hat{j}+\hat{k}$$

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

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

Solution

$$\overrightarrow{AB}=(4-1)\hat i +(3-1)\hat j+(1-2)\hat k=3\hat i+2\hat j -1\hat k$$
$$\overrightarrow{AC}=(2-1)\hat i +(3-1)\hat j+(5-2)\hat k=1\hat i+2\hat j+3\hat k$$
Bisector of two vectors is given by $$\overrightarrow{c}=||b||\overrightarrow{a}+||a||\overrightarrow{b}$$
$$||AB||=\sqrt{3^2+2^2+1^2}=\sqrt{9+4+1}=\sqrt{15}$$
$$||AC||=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{15}$$
Bisector of $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}=||AC||\overrightarrow{AB}+||AB||\overrightarrow{AC}$$
$$\Rightarrow\sqrt{15}(3\hat i+2\hat j -1\hat k)+\sqrt{15}(1\hat i+2\hat j+3\hat k)$$
$$\Rightarrow \sqrt{15}(4\hat i+4\hat j +2\hat k)$$
Simplifying the vector $$\dfrac{\sqrt{15}(4\hat i+4\hat j +2\hat k)}{\sqrt{15}\times 2}$$
$$\Rightarrow 2\hat i+2\hat j +1\hat k$$
Question 5
1 / -0

A unit vector along the direction $$\hat i + \hat j + \hat k$$ has a magnitude:

A
$$\sqrt 3 $$

B
$$\sqrt 2 $$

C
$$1$$

D
$$0$$

Solution

$$A\quad unit\quad vector\quad along\quad any\quad direction\quad always\quad has\quad magnitude=1$$
$$ C)Ans.$$
Question 6
1 / -0

Four forces act on a point object. The object will be in equilibrium, if:

A
all of them are in the same plane

B
they are opposite to each other in pairs

C
the sum of x, y and z - components of forces zero separately

D
they form a closed figure of 4 sides when added as Polygon law

Solution

the equilibrium condition is obtained when the net force acting on the body is zero. and a closed polygon of 4 sides will give the resultant force as zero and forcing acting will be in the same plane
Question 7
1 / -0

The vector that must be added to the vector $$\hat{i}-3\hat{j}+2\hat{k}$$ and $$3\hat{i}+6\hat{j}-7\hat{k}$$ so that the resultant vector is a unit vector along the y-axis is:

A
$$4\hat{i}+2\hat{j}+5\hat{k}$$

B
$$-4\hat{i}-2\hat{j}+5\hat{k}$$

C
$$3\hat{i}+4\hat{j}+5\hat{k}$$

D
Null vector

Solution

$$\textbf{Step 1: Equation formation as per given data}$$
Let vector $$\vec{C}$$ is added to vector $$\vec{A}$$ and $$\vec{B}$$ to give a resultant unit vector along $$y$$-axis.

$$\therefore \vec{A} + \vec{B} + \vec{C} = \hat{j}$$

$$\textbf{Step 2: Solving equation}$$
$$(\hat{i} - 3\hat{j} + 2 \hat{k}) + (3 \hat{i} + 6 \hat{j} - 7 \hat{k}) + \vec{C} = \hat{j}$$

$$\vec{C} = -4 \hat{i} - 2 \hat{j} + 5 \hat{k}$$

Hence the vector is $$\vec{C} = -4 \hat{i} - 2\hat{j} + 5 \hat{k}$$.
Option $$B$$ is correct.
Question 8
1 / -0

If $$\left| {\overrightarrow A \times \overrightarrow B | = \sqrt 3 \overrightarrow {A.} \overrightarrow B } \right.$$ then the value of $$\left| {\overrightarrow A + \overrightarrow B } \right|$$ is:

A
$$
\left( {A^2 + B^2 + \dfrac{{AB}}
{{\sqrt 3 }}} \right)^{1/2}
$$

B
A+ B

C
$$
\left( {A^2 + B^2 + \sqrt 3 AB} \right)^{1/2}
$$

D
$$
\left( {A^2 + B^2 + AB} \right)^{1/2}
$$

Solution

$${\textbf{Step -1: Find angle and use vector formula.}}$$
$${\text{We know that,}}$$
$$\Rightarrow \bar A× \bar B=ABsinθ$$
$${\text{Also,}}$$ $$ \bar A. \bar B=ABcosθ$$
$${\text{Given :}}$$ $$∣ \bar A× \bar B∣= \sqrt3A.B$$
$${\text{Using}}$$ $$∣ \bar A× \bar B∣=∣ \bar A∣∣ \bar B∣sinθ$$
$${\text{We get,}}$$ $$∣ \bar A× \bar B∣=ABsinθ$$
$$\Rightarrow \bar A.\bar B=∣ \bar A∣∣ \bar B∣cosθ$$
$$∴ ABsinθ= \sqrt3ABcosθ$$
$$\Rightarrow tanθ=\sqrt 3$$ $$⟹θ=60 ^\circ$$
$${\textbf{Step -2: Find }}$$ $${\mathbf{| \bar A + \bar B|.}}$$
$${\text{Now}}$$ $$(A+B)^2=A^2+B^2+2A.B$$
$$=A^2+B^2+2ABcosθ$$
$$=A^2+B^2+2AB× \dfrac{1}{2}$$
$$=A^2+B^2+AB$$
$${\text{or}}$$ $$∣ \bar A+\bar B∣=(A^2+B^2+AB)^{\frac{1}{2}}$$
$${\textbf{ Hence, the option (D) is correct.}}$$
Question 9
1 / -0

Which of the following is the unit vector perpendicular to $$ \vec{A} $$ and $$ \vec{B} $$ ?

A
$$ {{\vec{A}\times \vec{B}} \over {AB\sin \theta }}$$

B
$$\left|{{\vec{A}\times \vec{B}} \over {AB\sin \theta }}\right|$$

C
$$ {{\vec{A}\times \vec{B}} \over {AB\cos \theta }}$$

D
$$\left|{{\vec{A}\times \vec{B}} \over {AB\cos \theta }}\right|$$

Solution
Question 10
1 / -0

The vector equation of the plane containing the line $$\vec {r}=(-2\hat {i}-3\hat {j}+4\hat {k})+\lambda(3\hat {i}-2\hat {j}-\hat {k})$$ and the point $$\hat {i}+2\hat {j}+3\hat {k}$$ is:

A
$$\vec {r}\cdot(\hat {i}+3\hat {k})=10$$

B
$$\vec {r}\cdot(\hat {i}-3\hat {k})=10$$

C
$$\vec {r}\cdot(3\hat {i}+\hat {k})=10$$

D
$$none\ of\ these$$

Solution

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Vector Algebra Test - 51

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Vector Algebra Test - 51

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SHARING IS CARING

Question 1

$${\sin ^2}\frac{\theta }{2}$$

$${\cos ^2}\frac{\theta }{2}$$

$${\tan ^2}\frac{\theta }{2}$$

$${\cot ^2}\frac{\theta }{2}$$

Solution

Question 2

$${\pi}$$

$${\pi}\over{2}$$

zero

$${\pi}\over{4}$$

Solution

Question 3

$${{5\sqrt 3 } \over 3}\left( {\hat i + \hat j - \hat k} \right)$$

$${{5\sqrt 3 } \over 3}\left( {\hat i + \hat j + \hat k} \right)$$

$${{5\sqrt 3 } \over 3}\left( {\hat i - \hat j + \hat k} \right)$$

$${{5\sqrt 3 } \over 3}\left( {-\hat i + \hat j + \hat k} \right)$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$$\sqrt 3 $$

$$\sqrt 2 $$

$$1$$

$$0$$

Solution

Question 6

all of them are in the same plane

they are opposite to each other in pairs

the sum of x, y and z - components of forces zero separately

they form a closed figure of 4 sides when added as Polygon law

Solution

Question 7

$$4\hat{i}+2\hat{j}+5\hat{k}$$

$$-4\hat{i}-2\hat{j}+5\hat{k}$$

$$3\hat{i}+4\hat{j}+5\hat{k}$$

Null vector

Solution

Question 8

$$\left( {A^2 + B^2 + \dfrac{{AB}}{{\sqrt 3 }}} \right)^{1/2} $$

A+ B

$$\left( {A^2 + B^2 + \sqrt 3 AB} \right)^{1/2} $$

$$\left( {A^2 + B^2 + AB} \right)^{1/2} $$

Solution

Question 9

$$ {{\vec{A}\times \vec{B}} \over {AB\sin \theta }}$$

$$\left|{{\vec{A}\times \vec{B}} \over {AB\sin \theta }}\right|$$

$$ {{\vec{A}\times \vec{B}} \over {AB\cos \theta }}$$

$$\left|{{\vec{A}\times \vec{B}} \over {AB\cos \theta }}\right|$$

Solution

Question 10

$$\vec {r}\cdot(\hat {i}+3\hat {k})=10$$

$$\vec {r}\cdot(\hat {i}-3\hat {k})=10$$

$$\vec {r}\cdot(3\hat {i}+\hat {k})=10$$

$$none\ of\ these$$

Solution

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$$

{\pi}

$$

$$

{\pi}\over{2}

$$

$$

{\pi}\over{4}

$$

$$
\left( {A^2 + B^2 + \dfrac{{AB}}
{{\sqrt 3 }}} \right)^{1/2}
$$

$$
\left( {A^2 + B^2 + \sqrt 3 AB} \right)^{1/2}
$$

$$
\left( {A^2 + B^2 + AB} \right)^{1/2}
$$