Vector Algebra Test

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Vector Algebra Test - 47

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Question 1
1 / -0

The angle between the two vectors $$\hat { i } +\hat { j } +\hat { k }$$ and $$ 2\hat { i } -2\hat { j } +2\hat { k } $$ is equal to

A
$$\cos ^{ -1 }{ \left( \dfrac { 2 }{ 3 } \right) } $$

B
$$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 6 } \right) } $$

C
$$\cos ^{ -1 }{ \left( \dfrac { 5 }{ 6 } \right) } $$

D
$$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 18 } \right) } $$

E
$$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) } $$

Solution

$$\textbf{Step 1: Angle between two vectors }(\theta)$$
Dot product between the two vectors is defined as:
$$\vec{a}.\vec{b}= |\vec{a}||\vec{b}|\cos\theta$$
$$\Rightarrow \cos \theta = \dfrac {\vec {a} \cdot \vec {b}}{|\vec {a}| |\vec {b}|} = \dfrac {(\hat {i} + \hat {j} + \hat {k})\cdot (2\hat {i} - 2\hat {j} + 2\hat {k})}{\sqrt {3}\times \sqrt {4 + 4 + 4}}$$$$= \dfrac {2 - 2 + 2}{\sqrt {3} \times \sqrt {12}} = \dfrac {1}{3}$$
$$\Rightarrow\theta = \cos^{-1} \left (\dfrac {1}{3}\right )$$
Hence angle between vectors is $$\cos^{-1}\left (\dfrac {1}{3}\right )$$
Option (E) correct.
Question 2
1 / -0

Let $$u, v$$ and $$w$$ be vectors such that $$u + v + w = 0$$. If $$|u| = 3, |v| = 4$$ and $$|w| = 5$$, then $$u . v + v . w + w.u$$ is equal to

A
$$0$$

B
$$-25$$

C
$$25$$

D
$$50$$

E
$$47$$

Solution

Given, $$|u| = 3, |v| = 4 $$ and $$|w| = 5$$

Also, $$ u + v + w = 0$$

On squaring both sides, we get

$$|u|^2 + |v|^2 + |w|^2 + 2(u. v + v.w + w.u) = 0$$

$$\Rightarrow 3^2 + 4^2 + 5^2 + 2 (u.v + v.w + w.u) = 0$$

$$\Rightarrow 9 + 16 + 25 + 2(u.v + v.w + w.u) = 0$$

$$\Rightarrow u.v + v.w + w.u = - \dfrac{50}{2} = - 25$$
Question 3
1 / -0

If $$a=\hat { i } +\hat { j } +\hat { k } $$, $$b=4\hat { i } +3\hat { j } +4\hat { k }$$ and $$c=\hat { i } +\alpha \hat { j } +\beta \hat { k } $$ are coplanar and $$\left| c \right| =\sqrt { 3 } $$, then

A
$$\alpha =\sqrt { 2 } ,\beta =1$$

B
$$\alpha =1,\beta =\pm 1$$

C
$$\alpha =\pm 1,\beta =1$$

D
$$\alpha =\pm 1,\beta =-1$$

E
$$\alpha =-1,\beta =\pm 1$$

Solution

Given that, $$a=\hat { i } +\hat { j } +\hat { k } $$,
$$b=4\hat { i } +3\hat { j } +4\hat { k } $$
and $$c=\hat { i } +\alpha \hat { j } +\beta \hat { k }$$
Since, $$a, b$$ and $$c$$ are coplanar.
Then, $$\begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix}=0$$
$$\Rightarrow 1\left( 3\beta -4\alpha \right) -1\left( 4\beta -4 \right) +1\left( 4\alpha -3 \right) =0$$
$$\Rightarrow 3\beta -4\alpha -4\beta +4+4\alpha -3=0$$
$$\Rightarrow \beta =1$$
Also, $$\left| c \right| =\sqrt { 3 } $$
$$\Rightarrow \sqrt { 1+{ \alpha }^{ 2 }+{ \beta }^{ 2 } } =\sqrt { 3 }$$
$$\Rightarrow { \alpha }^{ 2 }=1\Rightarrow \alpha =\pm 1$$
Question 4
1 / -0

If $$\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k} , |\vec{b}| = 5 $$ and the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$ \dfrac{\pi}{6} $$, then the area of the triangle formed by these two vectors as two sides is

A
$$\dfrac{15}{4}$$

B
$$\dfrac{15}{2}$$

C
$$15$$

D
$$\dfrac{15\sqrt 3}{2}$$

E
$$15 \sqrt 3$$

Solution

Area of the triangle
$$=\dfrac{1}{2} |a \times b|$$
$$= \dfrac{1}{2} ||a| |b| \sin \theta \widehat n|$$
$$= \dfrac{1}{2} \left [ 3 \times 5 \times \sin \dfrac{\pi}{6} \right ]$$
$$\left( \therefore |a| = \sqrt{1 + 2^2 + 2^2 } = 3 \right )$$
$$= \dfrac{1}{2} \left [ 15 \times \dfrac{1}{2} \right ] = \dfrac{15}{4}$$
Question 5
1 / -0

If $$\lambda (3 \widehat i + 2 \widehat j - 6 \widehat k)$$ is a unit vector, then the values of $$\lambda $$ are

A
$$\pm \dfrac{1}{7}$$

B
$$\pm 7$$

C
$$\pm \sqrt{43}$$

D
$$\pm \dfrac{1}{\sqrt{43}}$$

E
$$\pm \dfrac{1}{\sqrt{}7}$$

Solution

Now, $$|(3 \widehat i + 2\widehat j - 6\widehat k)| = \sqrt{3^2 + 2^2 + (-6)^2}$$
$$= \sqrt{9 + 4 + 36}$$
$$= \sqrt{49} = 7$$
Since, $$\lambda (3 \widehat i + 2\widehat j - 6\widehat k)$$ is a unit vector.
$$\therefore \lambda \pm \dfrac{1}{|3 \widehat i + 2\widehat j - 6\widehat k|}$$
$$= \pm \dfrac{1}{7}$$
Question 6
1 / -0

Let $$a,b,c$$ be three non-zero vectors such that $$a+b+c=0$$, then $$\lambda b\times a+b\times c+c\times a=0$$, where $$\lambda$$ is

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$-2$$

Solution

Clearly $$a,b,c$$ represents the sides of a triangle.
Therefore, $$ a\times b=b\times c=c\times a\quad $$
$$ \Rightarrow a\times b+b\times c+c\times a=3a\times b$$
$$\Rightarrow 2b\times a+b\times c+c\times a=0$$
$$\Rightarrow \lambda =2$$
Question 7
1 / -0

Let $$ABCD$$ be a parallelogram. If $$AB=\hat { i } +3\hat { j } +7\hat { k } , AD=2\hat { i } +3\hat { j } -5\hat { k } $$ and $$p$$ is a unit vector parallel to $$AC$$, then $$p$$ is equal to

A
$$\dfrac { 1 }{ 3 } \left( 2\hat { i } +\hat { j } +2\hat { k } \right) $$

B
$$\dfrac { 1 }{ 3 } \left( 2\hat { i } +2\hat { j } +2\hat { k } \right) $$

C
$$\dfrac { 1 }{ 7 } \left( 3\hat { i } +6\hat { j } +2\hat { k } \right) $$

D
$$\dfrac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } +3\hat { k } \right) $$

E
$$\dfrac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right) $$

Solution

Given, $$AB=\hat i+3\hat j+7\hat k, AD=2\hat i+3\hat j-5\hat k$$
$$AC=AB+AD$$
$$=\left( \hat { i } +3\hat { j } +7\hat { k } \right) +\left( 2\hat { i } +3\hat { j } -5\hat { k } \right) $$
$$=3\hat { i } +6\hat { j } +2\hat { k } $$
Now, $$p$$ is a unit vector parallel to $$AC$$.
Then, $$p=\dfrac { AC }{ \left| AC \right| } $$
$$=\dfrac { 3\hat { i } +6\hat { j } +2\hat { k } }{ \sqrt { 9+36+4 } }$$
$$\Rightarrow p=\dfrac { 1 }{ 7 } \left( 3\hat { i } +6\hat { j } +2\hat { k } \right) $$
Question 8
1 / -0

If $$a=2i+2j+3k, b=-1+2j+k$$ and $$c = 3i + j$$, then $$a+tb$$ this perpendicular to $$c$$; if $$t$$ is equal to

A
2

B
4

C
6

D
8

Solution

Given, $$a=2i+2j+3k,b=-1+2j+k $$ and $$c=3i+j$$
We need to find value of $$a+ tb\perp c$$
$$\Rightarrow (a+tb)\cdot c=0$$
$$\Rightarrow a\cdot c+ tb \cdot c=0$$
$$\Rightarrow t=-\dfrac {a\cdot c}{b\cdot c}$$
$$\Rightarrow t=-\dfrac {(2i+2j+3k)\cdot (3i+j)}{(-i+2j+k)\cdot (3i+j)}$$
$$\Rightarrow t=-\dfrac {6+2+0}{-3+2+0}=8$$
Question 9
1 / -0

Let $$P\left( 1,2,3 \right) $$ and $$Q\left( -1,-2,-3 \right) $$ be the two points and let $$O$$ be the origin. Then, $$\left| PQ+OP \right| $$ is equal to

A
$$\sqrt { 13 } $$

B
$$\sqrt { 14 } $$

C
$$\sqrt { 24 } $$

D
$$\sqrt { 12 } $$

E
$$\sqrt { 8 } $$

Solution

Given that, $$P=\left( 1,2,3 \right) , Q=\left( -1,-2,-3 \right) $$ and $$O=\left( 0,0,0 \right) $$.
Then, $$PQ=\left( -1-1 \right) \hat { i } +\left( -2-2 \right) \hat { j } +\left( -3-3 \right) \hat { k } $$
$$=-2\hat { i } -4\hat { j } -6\hat { k } $$
and $$OP=\left( 1-0 \right) \hat { i } +\left( 2-0 \right) \hat { j } +\left( 3-0 \right) \hat { k } $$
$$=\hat { i } +2\hat { j } +3\hat { k } $$
Now, $$PQ+OP=\left( -2\hat { i } -4\hat { j } -6\hat { k } \right) +\left( \hat { i } +2\hat { j } +3\hat { k } \right) $$
$$=-\hat { i } -2\hat { j } -3\hat { k } $$
$$\Rightarrow \left| PQ+OP \right| =\sqrt { { \left( -1 \right) }^{ 2 }+{ \left( -2 \right) }^{ 2 }+{ \left( -3 \right) }^{ 2 } } =\sqrt { 14 } $$

Alternate Method
$$PQ=OQ-OP$$
$$\Rightarrow PQ+OP=OQ$$
Now, $$OQ=-\hat { i } -2\hat { j } -3\hat { k } $$
$$\left| OQ \right| =\sqrt { { \left( -1 \right) }^{ 2 }+{ \left( -2 \right) }^{ 2 }+{ \left( -3 \right) }^{ 2 } } $$ $$=\sqrt { 14 } $$
Therefore, $$ \left| PQ+OP \right| =\left| OQ \right| $$ $$=\sqrt { 14 } $$
Question 10
1 / -0

If $$3p + 2q =i + j + k$$ and $$3p - 2q = i - j - k$$, then the angle between $$p$$ and $$q$$ is

A
$$\dfrac {\pi}{6}$$

B
$$\dfrac {\pi}{4}$$

C
$$\dfrac {\pi}{3}$$

D
$$\dfrac {\pi}{2}$$

E
$$\pi$$

Solution

Given, $$3p + 2q = i + j + k $$ ..... (i)
and $$3p - 2q = i - j - k$$ .... (ii)
On adding Eqs. (i) and (ii), we get
$$6p = 2i$$
Therefore, $$ p = \dfrac {i}{3}$$
On subtracting Eq. (ii) from Eq. (i), we get
$$4q - 2(j + k) \Rightarrow q = \dfrac {1}{2} (j + k)$$
Let $$\theta$$ be the angle between $$p$$ and $$q$$, then
$$p\cdot q = |p||q|\cos \theta$$
$$\Rightarrow \cos \theta = \dfrac {p\cdot q}{|p||q|}$$
$$\Rightarrow \cos \theta = \dfrac {1}{6} \dfrac {i\cdot (j + k)}{|p||q|} = 0 = \cos 90^{\circ}$$
Thus $$ \theta = 90^{\circ} = \dfrac {\pi}{2}$$

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Vector Algebra Test - 47

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Vector Algebra Test - 47

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SHARING IS CARING

Question 1

$$\cos ^{ -1 }{ \left( \dfrac { 2 }{ 3 } \right) } $$

$$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 6 } \right) } $$

$$\cos ^{ -1 }{ \left( \dfrac { 5 }{ 6 } \right) } $$

$$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 18 } \right) } $$

$$\cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 } \right) } $$

Solution

Question 2

$$0$$

$$-25$$

$$25$$

$$50$$

$$47$$

Solution

Question 3

$$\alpha =\sqrt { 2 } ,\beta =1$$

$$\alpha =1,\beta =\pm 1$$

$$\alpha =\pm 1,\beta =1$$

$$\alpha =\pm 1,\beta =-1$$

$$\alpha =-1,\beta =\pm 1$$

Solution

Question 4

$$\dfrac{15}{4}$$

$$\dfrac{15}{2}$$

$$15$$

$$\dfrac{15\sqrt 3}{2}$$

$$15 \sqrt 3$$

Solution

Question 5

$$\pm \dfrac{1}{7}$$

$$\pm 7$$

$$\pm \sqrt{43}$$

$$\pm \dfrac{1}{\sqrt{43}}$$

$$\pm \dfrac{1}{\sqrt{}7}$$

Solution

Question 6

$$1$$

$$2$$

$$-1$$

$$-2$$

Solution

Question 7

$$\dfrac { 1 }{ 3 } \left( 2\hat { i } +\hat { j } +2\hat { k } \right) $$

$$\dfrac { 1 }{ 3 } \left( 2\hat { i } +2\hat { j } +2\hat { k } \right) $$

$$\dfrac { 1 }{ 7 } \left( 3\hat { i } +6\hat { j } +2\hat { k } \right) $$

$$\dfrac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } +3\hat { k } \right) $$

$$\dfrac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right) $$

Solution

Question 8

2

4

6

8

Solution

Question 9

$$\sqrt { 13 } $$

$$\sqrt { 14 } $$

$$\sqrt { 24 } $$

$$\sqrt { 12 } $$

$$\sqrt { 8 } $$

Solution

Question 10

$$\dfrac {\pi}{6}$$

$$\dfrac {\pi}{4}$$

$$\dfrac {\pi}{3}$$

$$\dfrac {\pi}{2}$$

$$\pi$$

Solution

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