Vector Algebra Test

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Vector Algebra Test - 64

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Question 1
1 / -0

If $$a, b, c$$ are vectors such that $$a+b+c = 0$$ and $$|a| = 7, |b| = 5, |c| = 3$$, then the angle between $$c$$ and $$b$$ is

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

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

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

D
$$\pi$$

Solution

We have,

$$a+b+c = 0$$
$$\Rightarrow b+c = -a$$
$$\Rightarrow |b+c| = |-a|$$

$$\Rightarrow |b+c| = |a|$$
$$\Rightarrow |b+c|^2 = |a|^2$$

$$\Rightarrow (b+c) . (b+c) = |a|^2$$
$$\Rightarrow |b|^2+|c|^2+2|b||c| \cos \theta = |a|^2$$

$$\Rightarrow (5)^2 + (3)^2 + 2\times 5\times 3 \cos \theta = (7)^2$$
$$\Rightarrow 25 + 9 + 30 \cos \theta = 49$$

$$\Rightarrow 30 \cos \theta = 15$$
$$\Rightarrow \cos \theta = \dfrac{1}{2}$$

$$\Rightarrow \theta = 60^o$$ or $$\dfrac{\pi}{3}$$

$$\therefore$$ Angle betweenn $$b$$ and $$c$$ is $$\dfrac{\pi}{3}$$.
Question 2
1 / -0

$$(\vec r. \hat i)(\vec r \times \hat i)+ (\vec r. \hat j)(\vec r \times \hat j) +(\vec r. \hat k)(\vec r \times \hat k)$$ is equal to

A
$$3\ \vec r$$

B
$$\vec r$$

C
$$\vec 0$$

D
$$None\ of\ these$$

Solution
Question 3
1 / -0

If $$(\vec a\times \vec b)^2 +(\vec a. \vec b)^2 =144$$ and $$|\vec a|=4$$, then $$|\vec b|=$$

A
$$16$$

B
$$8$$

C
$$3$$

D
$$12$$

Solution
Question 4
1 / -0

$$p\hat{i}+3\hat{j}+4\hat{k}$$ and $$\sqrt{q}\hat{i}+4\hat{k}$$ are two vectors, where $$p,q>0$$ are two scalars, then the length of the vectors is equal to

A
All value of $$(p,q)$$

B
Only finite number of values of $$(p,q)$$

C
Infinite number of values of $$(p,q)$$

D
No value fo $$(p,q)$$

Solution
Question 5
1 / -0

Given that $$\vec a, \vec b, \vec p, \vec q$$ are four vectors such that $$\vec a + \vec b = \mu \vec p, \vec b \cdot \vec q = 0$$ and $$(\vec b)^2 = 1,$$ where $$\mu$$ is scalar. Then $$\mid (\vec a \cdot \vec q) \vec p - (\vec p \cdot \vec q)\vec a \mid$$ is equal to

A
$$2 \mid \vec p \cdot \vec q \mid$$

B
$$(1/2) \mid \vec p \cdot \vec q \mid$$

C
$$\mid \vec p \times \vec q \mid$$

D
$$\mid \vec p \cdot \vec q \mid$$

Solution

$$\vec a + \vec b = \mu \vec p,$$ $$ \vec b \cdot \vec q = 0$$, $$(\vec b)^2 = 1,$$

$$\because \vec a + \vec b = \mu \vec p$$

$$\Rightarrow (\vec a + \vec b) \times \vec a = \mu \vec p \times \vec a, \vec b \times \vec a = \mu \vec p \times \vec a \Rightarrow \vec q \times (\vec b \times \vec a) = \mu \vec q \times (\vec p \times \vec q)$$

$$\Rightarrow (\vec q \cdot \vec a) \vec b - (\vec q \cdot \vec b) \vec a = \mu \vec q \times (\vec p \times \vec a) \Rightarrow (\vec q \cdot \vec a) \vec b = \mu \vec q \times (\vec p \times \vec a)$$

$$\because \vec a + \vec b = \mu \vec p$$

$$\Rightarrow \vec q \cdot (\vec a + \vec b) = \mu \vec q \cdot \vec p$$

$$\Rightarrow \vec q \cdot \vec a + \vec q \cdot \vec b = \mu \vec p \cdot \vec q$$

$$\mu = \dfrac{\vec q \cdot \vec a}{\vec p \cdot \vec q}$$

$$\Rightarrow (\vec q \cdot \vec a) \vec b = \dfrac{\vec q \cdot \vec a}{\vec p \cdot \vec q} [(\vec q \cdot \vec a) \cdot \vec p - (\vec q \cdot \vec p)\vec a]$$

$$\Rightarrow \mid (\vec q \cdot \vec a)\vec p - (\vec q \cdot \vec p)\vec a \mid = \mid (\vec p \cdot \vec q) \vec b \mid = \mid (\vec p \cdot \vec q)\mid \cdot \mid \vec b\mid$$

$$\Rightarrow \mid (\vec q \cdot \vec a)\vec p - (\vec q \cdot \vec p) \vec a \mid = \mid \vec p \cdot \vec q \mid$$
Question 6
1 / -0

Directions For Questions

If $$ \vec{x} \times \vec{y}=\vec{a}, \vec{y} \times \vec{z}=\vec{b}, \vec{x} \cdot \vec{b}=\gamma, \vec{x} \cdot \vec{y}=1 $$ and $$ \vec{y} \cdot \vec{z}=1 $$

...view full instructions

Vector $$ \vec{x} $$ is

A
$$\dfrac{1}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times(\vec{a} \times \vec{b})]$$

B
$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$$

C
$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$$

D
none of these

Solution

Given
$$ \vec{x} \times \vec{y}=\vec{a} $$ $$\qquad (i)$$
$$ \vec{y} \times \vec{z}=\vec{b} $$ $$\qquad (ii)$$
$$ \vec{x} \cdot \vec{b}=\gamma $$ $$\qquad (iii)$$
$$ \vec{x} \cdot \vec{y}=1 $$ $$\qquad (iv)$$
$$ \vec{y} \cdot \vec{z}=1 $$ $$\qquad (v)$$
From (ii), $$ \vec{x} \cdot(\vec{y} \times \vec{z})=\vec{x} \cdot \vec{b}=\gamma \Rightarrow[\vec{x} \vec{y} \vec{z}]=\gamma $$
From (i) and (ii), $$ (\vec{x} \times \vec{y}) \times(\vec{y} \times \vec{z})=\vec{a} \times \vec{b} $$
$$ \therefore[\vec{x} \vec{y} \vec{z}] \vec{y}-[\vec{y} \vec{y} \vec{z}] \vec{x}=\vec{a} \times \vec{b} \Rightarrow \vec{y}=\dfrac{\vec{a} \times \vec{b}}{\gamma} $$ $$\qquad (vi)$$
Also from (i), we get $$ (\vec{x} \times \vec{y}) \times \vec{y}=\vec{a} \times \vec{y} $$
$$ \Rightarrow(\vec{x} \cdot \vec{y}) \vec{y}-(\vec{y} \cdot \vec{y}) \vec{x}=\vec{a} \times \vec{y} \Rightarrow \vec{x}=\left(1 /|\vec{y}|^{2}\right)(\vec{y}-\vec{a} \times \vec{y})=\dfrac{\gamma^{2}}{|\vec{a} \times \vec{b}|^{2}}\left[\dfrac{\vec{a} \times \vec{b}}{\gamma}-\dfrac{\vec{a} \times(\vec{a} \times \vec{b})}{\gamma}\right] $$
$$\begin{array}{l}\Rightarrow \vec{x}=\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}-\vec{a} \times(\vec{a} \times \vec{b})] \end{array}$$
Question 7
1 / -0

Directions For Questions

If $$ \vec{x} \times \vec{y}=\vec{a}, \vec{y} \times \vec{z}=\vec{b}, \vec{x} \cdot \vec{b}=\gamma, \vec{x} \cdot \vec{y}=1 $$ and $$ \vec{y} \cdot \vec{z}=1 $$

...view full instructions

Vector $$ \vec{z} $$ is

A
$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b} \times(\vec{a} \times \vec{b})] $$

B
$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$$

C
$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$$

D
none of these

Solution

Given
$$ \vec{x} \times \vec{y}=\vec{a} $$ $$\qquad (i)$$
$$ \vec{y} \times \vec{z}=\vec{b} $$ $$\qquad (ii)$$
$$ \vec{x} \cdot \vec{b}=\gamma $$ $$\qquad (iii)$$
$$ \vec{x} \cdot \vec{y}=1 $$ $$\qquad (iv)$$
$$ \vec{y} \cdot \vec{z}=1 $$ $$\qquad (v)$$

$$\begin{array}{l} \text { Also from (ii), }(\vec{y} \times \vec{z}) \times \vec{y}=\vec{b} \times \vec{y} \Rightarrow|\vec{y}|^{2} \vec{z}-(\vec{z} \cdot \vec{y}) \vec{y}=\vec{b} \times \vec{y} \\\Rightarrow \vec{z}=\dfrac{1}{|\vec{y}|^{2}}[\vec{y}+\vec{b} \times \vec{y}]=\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]\end{array}$$
Question 8
1 / -0

Directions For Questions

If $$ \vec{x} \times \vec{y}=\vec{a}, \vec{y} \times \vec{z}=\vec{b}, \vec{x} \cdot \vec{b}=\gamma, \vec{x} \cdot \vec{y}=1 $$ and $$ \vec{y} \cdot \vec{z}=1 $$

...view full instructions

Vector $$ \vec{y} $$ is

A
$$\dfrac{\vec{a} \times \vec{b}}{\gamma}$$

B
$$\vec{a}+\dfrac{\vec{a} \times \vec{b}}{\gamma}$$

C
$$\vec{a}+\vec{b}+\dfrac{\vec{a} \times \vec{b}}{\gamma} $$

D
none of these

Solution

Given
$$ \vec{x} \times \vec{y}=\vec{a} $$ $$\qquad (i)$$
$$ \vec{y} \times \vec{z}=\vec{b} $$ $$\qquad (ii)$$
$$ \vec{x} \cdot \vec{b}=\gamma $$ $$\qquad (iii)$$
$$ \vec{x} \cdot \vec{y}=1 $$ $$\qquad (iv)$$
$$ \vec{y} \cdot \vec{z}=1 $$ $$\qquad (v)$$
From (ii), $$ \vec{x} \cdot(\vec{y} \times \vec{z})=\vec{x} \cdot \vec{b}=\gamma \Rightarrow[\vec{x} \vec{y} \vec{z}]=\gamma $$
From (i) and (ii), $$ (\vec{x} \times \vec{y}) \times(\vec{y} \times \vec{z})=\vec{a} \times \vec{b} $$
$$ \therefore[\vec{x} \vec{y} \vec{z}] \vec{y}-[\vec{y} \vec{y} \vec{z}] \vec{x}=\vec{a} \times \vec{b} \Rightarrow \vec{y}=\dfrac{\vec{a} \times \vec{b}}{\gamma} $$
Question 9
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If $$\vec r$$ and $$\vec s$$ are non-zero constant vectors and the scalar $$b$$ is chosen such that $$\mid \vec r + b \vec s \mid$$ is minimum, then the value of $$\mid b \vec s \mid^2 + \mid \vec r + b \vec s \mid^2$$ is equal to

A
$$2 \mid \vec r \mid^2$$

B
$$\mid \vec r \mid^2/2$$

C
$$3 \mid \vec r \mid^2$$

D
$$\mid \vec r \mid^2$$

Solution

For minimum value $$\mid \vec r + b \vec s \mid = 0$$
Let $$\vec r$$ and $$\vec s$$ are anti parallel so $$b \vec s = - \vec r$$
so $$\mid b \vec s\mid^2 + \mid \vec r + b \vec s \mid^2 = \mid - \vec r \mid^2 + \mid \vec r - \vec r \mid^2 = \mid \vec r \mid^2$$
Question 10
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Directions For Questions

Given two orthogonal vectors $$ \vec{A} $$ and $$ \vec{B} $$ each of length unity. Let $$ \vec{P} $$ be the vector satisfying the equation $$ \vec{P} \times \vec{B}=\vec{A}-\vec{P} \cdot $$ Then

...view full instructions

$$ (\vec{P} \times \vec{B}) \times \vec{B} $$ is equal to

A
$$\vec{P}$$

B
$$-\vec{P}$$

C
$$2 \vec{B}$$

D
$$\vec{A}$$

Solution

$$ \vec{P} \times \vec{B}=\vec{A}-\vec{P} $$ and $$ |\vec{A}|=|\vec{B}|=1 $$ and $$ \vec{A} \cdot \vec{B}=0 $$ is given
Now $$ \vec{P} \times \vec{B}=\vec{A}-\vec{P} $$ $$ \qquad (i)$$
$$ (\vec{P} \times \vec{B}) \times \vec{B}=(\vec{A}-\vec{P}) \times \vec{B} $$ (taking cross product with $$ \vec{B} $$ on both sides)
$$\begin{array}{l}\Rightarrow(\vec{P} \cdot \vec{B}) \vec{B}-(\vec{B} \cdot \vec{B}) \vec{P}=\vec{A} \times \vec{B}-\vec{P} \times \vec{B} \\\Rightarrow(\vec{P} \cdot \vec{B}) \vec{B}-\vec{P}=\vec{A} \times \vec{B}-\vec{A}+\vec{P} \\\Rightarrow 2 \vec{P}=\vec{A}-\vec{A} \times \vec{B}-(\vec{P} \cdot \vec{B}) \vec{B} \\\Rightarrow \vec{P}=\dfrac{\vec{A}-\vec{A} \times \vec{B}-(\vec{P} \cdot \vec{B}) \vec{B}}{2}\end{array}$$ $$ \qquad (ii)$$

Taking dot product with $$ \vec{B} $$ on both sides of (i), we get
$$\begin{array}{l}\vec{P} \cdot \vec{B}=\vec{A} \cdot \vec{B}-\vec{P} \cdot \vec{B} \\\Rightarrow \vec{P} \cdot \vec{B}=0 \qquad \qquad (iii) \\\Rightarrow \vec{P}=\dfrac{\vec{A}+\vec{B} \times \vec{A}}{2}\end{array}$$

Now
$$(\vec{P} \times \vec{B}) \times \vec{B}=(\vec{P} \cdot \vec{B}) \vec{B}-(\vec{B} \cdot \vec{B}) \vec{P}=-\vec{P}$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 64

Result

Vector Algebra Test - 64

Score

-

Rank

-

SHARING IS CARING

Question 1

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

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

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

$$\pi$$

Solution

Question 2

$$3\ \vec r$$

$$\vec r$$

$$\vec 0$$

$$None\ of\ these$$

Solution

Question 3

$$16$$

$$8$$

$$3$$

$$12$$

Solution

Question 4

All value of $$(p,q)$$

Only finite number of values of $$(p,q)$$

Infinite number of values of $$(p,q)$$

No value fo $$(p,q)$$

Solution

Question 5

$$2 \mid \vec p \cdot \vec q \mid$$

$$(1/2) \mid \vec p \cdot \vec q \mid$$

$$\mid \vec p \times \vec q \mid$$

$$\mid \vec p \cdot \vec q \mid$$

Solution

Question 6

$$\dfrac{1}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times(\vec{a} \times \vec{b})]$$

$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$$

$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$$

none of these

Solution

Question 7

$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b} \times(\vec{a} \times \vec{b})] $$

$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a}+\vec{b}-\vec{a} \times(\vec{a} \times \vec{b})]$$

$$\dfrac{\gamma}{|\vec{a} \times \vec{b}|^{2}}[\vec{a} \times \vec{b}+\vec{b} \times(\vec{a} \times \vec{b})]$$

none of these

Solution

Question 8

$$\dfrac{\vec{a} \times \vec{b}}{\gamma}$$

$$\vec{a}+\dfrac{\vec{a} \times \vec{b}}{\gamma}$$

$$\vec{a}+\vec{b}+\dfrac{\vec{a} \times \vec{b}}{\gamma} $$

none of these

Solution

Question 9

$$2 \mid \vec r \mid^2$$

$$\mid \vec r \mid^2/2$$

$$3 \mid \vec r \mid^2$$

$$\mid \vec r \mid^2$$

Solution

Question 10

$$\vec{P}$$

$$-\vec{P}$$

$$2 \vec{B}$$

$$\vec{A}$$

Solution

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