Vector Algebra Test

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Vector Algebra Test - 50

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Question 1
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Let $$\vec {a} = x\hat {i} + 12\hat {j} - \hat {k}, \vec {b} = 2\hat {i} + 2x\hat {j} + \hat {k}$$ and $$\vec {c} = \hat {i} + \hat {k}$$. If ordered set $$[\vec {b} \vec {c} \vec {a}]$$ is left handed, then.

A
$$x\epsilon (2, \infty)$$

B
$$x\epsilon (-\infty, -3)$$

C
$$x\epsilon (-3, 2)$$

D
$$x\epsilon \left \{3, 2\right \}$$

Solution

Given,
$$\vec { a } =x\hat { i } +12\hat { j } -\hat { k } \\ \vec { b } =2\hat { i } +2x\hat { j } +\hat { k } \\ \vec { c } =\hat { i } +\hat { k } $$
the ordered set $$[\vec { b } \vec { c } \vec { a } ]$$ is left handed if and only if
$$[\vec { b } \vec { c } \vec { a } ]<0$$
Now,
$$[\vec { b } \vec { c } \vec { a } ]=\vec { b } .(\vec { c } \times \vec { d } )=2{ x }^{ 2 }+2x-12=(2x-4)(x+3)$$
the expression $$2{ x }^{ 2 }+2x-12$$ is negative when $$x\quad \epsilon (-3,2)$$
Hence $$[\vec { b } \vec { c } \vec { a } ]$$ is left handed for $$x\quad \epsilon (-3,2)$$
Question 2
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Which is a unit vector?

A
$$(Cos \alpha, 2 Sin \alpha)$$

B
$$(Sin \alpha, Cos \alpha)$$

C
$$(1, -1)$$

D
$$(2Cos \alpha, Sin \alpha)$$

Solution

$$(\sin \alpha/\cos\alpha)$$
Let $$\vec{a}=\sin\alpha\hat{i}$$$$+\cos\alpha\hat{j}\implies |\vec{x}|=\sqrt{\sin^2\alpha+\cos^2\alpha}=1$$
$$\therefore (\sin\alpha,\cos\alpha)$$ is a unit vector.
Question 3
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If $$\vec {a}$$ and $$\vec {b}$$ are unit vectors, then angle between $$\vec {a}$$ and $$\vec {b}$$ for $$\sqrt {3} \vec {a} - \vec {b}$$ to be unit vector is

A
$$60^{\circ}$$

B
$$90^{\circ}$$

C
$$45^{\circ}$$

D
$$30^{\circ}$$

Solution

$$\textbf{Step 1: Dot product of vectors}$$
Let angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\theta$$
Dot product between them:
$$\vec{a}\cdot \vec{b} = |\vec{a}| |\vec{b}|\cos \theta$$
As $$\vec{a}$$ and $$\vec{b}$$ are unit vetors therefore $$|\vec{a}|=|\vec{b}|=1$$
$$\Rightarrow \vec{a}\cdot \vec{b} = \cos \theta$$ $$....(1)$$

$$\textbf{Step 2: Equation formation}$$
For $$\sqrt{3}\vec{a}-\vec{b}$$ to be a unit vector,
$$|\sqrt{3}\vec{a} - \vec{b}|^2 = 1$$

$$\Rightarrow (\sqrt{3})^2 |\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{3} \vec{a}\cdot \vec{b} = 1$$

$$\Rightarrow (\sqrt{3})^2 (1)^2 + (1)^2 - 2\sqrt{3} \cos \theta=1$$ [ Using $$(1)$$]

$$\Rightarrow 4 - 2\sqrt{3} \cos\theta=1$$

$$\Rightarrow$$ $$\dfrac{-3}{-2\sqrt{3}} = \cos \theta$$

$$\Rightarrow$$ $$\cos \theta = \dfrac{\sqrt{3}}{2}$$

$$\Rightarrow$$ $$\theta = 30^o$$
Hence option $$D$$ is correct.
Question 4
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If $$ \vec a+2\vec b+3 \vec c =0$$, then $$\vec{b}\times \vec{c}+\vec{c}\times \vec{a}+\vec{a} \times \vec{b}$$ equals to:

A
$$6\left( b\times c \right)$$

B
$$\left( a\times b \right)$$

C
$$6\left( c\times a \right)$$

D
0

Solution

$$\vec a+2{\vec b}+3{\vec c}=0$$
$$(\vec a+2\vec b+3\vec c)\times \vec c=0\implies \vec c\times \vec a=2(\vec b\times \vec c)$$
$$(\vec a+2\vec b+3\vec c)\times \vec b=0\implies \vec a\times \vec b=3(\vec b\times \vec c)$$
$$\vec b\times \vec c+\vec c\times \vec a+\vec a\times \vec b=(\vec b\times \vec c)+2(\vec b\times \vec c)+3(\vec b\times \vec c)=6(\vec b\times \vec c)=3(\vec c\times \vec a)=2(\vec a\times \vec b)$$
Question 5
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Let $$\overline a ,\overline b ,\overline c $$ be vectors of length $$3, 4, 5 $$respectively. Let $$\overline a $$ be perpendicular to $$\overline b + \overline c ,\overline b \,to\,\overline c + \overline a \,{\text{and}}\,\overline c \,{\text{to}}\,\overline a + \overline b .\,{\text{Then}}|\overline a + \overline b + \overline c |$$ is equals to:

A
$$2\sqrt 5 $$

B
$$2\sqrt 2 $$

C
$$10\sqrt 5 $$

D
$$5\sqrt 2 $$

Solution

According to question,
$$\vec { a } .\left( \vec { b } +\vec { c } \right) =0$$
$$\vec { b } .\left( \vec { c } +\vec { a } \right) =0$$
$$\vec { c } .\left( \vec { a } +\vec { b } \right) =0$$
$$\Rightarrow \quad \vec { a } .\vec { b } +\vec { a } .\vec { c } =0\quad \quad \quad \rightarrow \left( 1 \right) $$
$$\Rightarrow \quad \vec { b } .\vec { c } +\vec { b } .\vec { a } =0\quad \quad \quad \rightarrow \left( 2 \right) $$
$$\Rightarrow \quad \vec { c } .\vec { a } +\vec { c } .\vec { b } =0\quad \quad \quad \rightarrow \left( 3 \right) $$
$${ \left| \vec { a } +\vec { b } +\vec { c } \right| }^{ 2 }={ \left| a \right| }^{ 2 }+{ \left| b \right| }^{ 2 }+{ \left| c \right| }^{ 2 }+2\left( \vec { a } .\vec { b } +\vec { b } .\vec { c } +\vec { a } .\vec { c } \right) \quad \quad \quad \rightarrow (4) $$
Adding (1), (2) and (3),
$$2\left( \vec { a } .\vec { b } +\vec { b } .\vec { c } +\vec { a } .\vec { c } \right) =0\quad \quad \quad (\because \vec { a } .\vec { b } =\vec { b } .\vec { a } )$$
$$\Rightarrow { \left| \vec { a } +\vec { b } +\vec { c } \right| }^{ 2 }={ \left| a \right| }^{ 2 }+{ \left| b \right| }^{ 2 }+{ \left| c \right| }^{ 2 }+0$$
$$\Rightarrow \quad { \left| \vec { a } +\vec { b } +\vec { c } \right| }^{ 2 }=9+16+25=50$$
$$\Rightarrow \quad { \left| \vec { a } +\vec { b } +\vec { c } \right| }=5\sqrt { 2 } $$
Question 6
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If the projection of $$\vec{a}$$ on $$\vec{b}$$ and the projection of $$\vec{b}$$ on $$\vec{a}$$ are equal then the angle between $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ is

A
$$\large{\frac{\pi}{3}}$$

B
$$\large{\frac{\pi}{2}}$$

C
$$\large{\frac{\pi}{4}}$$

D
$$\large{\frac{2\pi}{3}}$$

Solution

Projection of $$\vec{a}$$ on $$\vec{b}$$ $$=\large{\frac{\vec{a}.\vec{b}}{|\vec{b}|}}$$, and projection of $$\vec{b}$$ on $$\vec{a}$$ $$=\large{\frac{\vec{a}.\vec{b}}{|\vec{a}|}}$$
$$\therefore \large{\frac{\vec{a}.\vec{b}}{|\vec{b}|}}=\large{\frac{\vec{a}.\vec{b}}{|\vec{a}|}}$$
$$\therefore |\vec{b}|=|\vec{a}|$$.
Now, angle between $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ be $$\theta$$.
Then $$\cos \theta=\large{\frac{(\vec{a}+\vec{b}).(\vec{a}-\vec{b})}{|\vec{a}+\vec{b}||\vec{a}-\vec{b}|}}$$
or, $$\cos \theta =\large{\frac{|\vec{a}|^2-|\vec{b}|^2}{|\vec{a}+\vec{b}||\vec{a}-\vec{b}|}}$$
or, $$\cos \theta =0$$ [Since, $$ |\vec{b}|=|\vec{a}|$$]
or, $$\theta=\large{\frac{\pi}{2}}$$
Question 7
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If $$\vec a$$ is perpendicular to $$\vec b$$ and $$\vec c,$$ then

A
$$\vec a \times \left( {\vec b \times \vec c} \right) = 1$$

B
$$\vec a \times \left( {\vec b \times \vec c} \right) = 0$$

C
$$\vec a \times \left( {\vec b \times \vec c} \right) = -1$$

D
None of the above

Solution

If $$\vec { a } $$ is perpendicular to both $$\vec { b } \quad and\quad \vec { c } $$
Let $$\vec { a } =\hat { i } $$
If $$\vec { b } =2\hat { j } $$
and $$\vec { c } =4\hat { j } $$
$$\therefore \quad \vec { a } \quad is\quad \bot \quad both\quad \vec { b } \quad and\quad \vec { c } $$
and $$\quad \vec { b } \times \vec { c } =\vec { 0 } $$ (null vector)
$$\Rightarrow \quad \vec { a } \times \left( \vec { b } \times \vec { c } \right) \quad =\vec { 0 } $$
Similarly, If $$\vec { b } =4\hat { k } $$
$$\vec { c } =5\hat { k } $$
$$\Rightarrow \quad \vec { a } \times \left( \vec { b } \times \vec { c } \right) \quad =\vec { 0 } $$
Now, If $$\vec { b } =5\hat { j } $$
$$\vec { c } =9\hat { k } $$
$$\Rightarrow \quad \vec { a } \times \left( \vec { b } \times \vec { c } \right) \quad =\vec { 0 } $$
($$\because \quad \vec { b } \times \vec { c } $$ results in a vector along $$\hat { i } $$ )
On R.H.S, $$0$$ is scalar is given and cross product of two vectors can never be scalar, hence it is actually null vector.
Question 8
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Let $$\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \,\,\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$$ and $$\vec{c} = \vec{i} + \vec{j} - 2\vec{k} $$ be three vectors. A vector of the type $$\vec{b} + \lambda \vec{c}$$ for some scalar $$\lambda$$, whose projection on $$\vec{a}$$ is of magnitude $$\sqrt {\frac{2}{3}}$$. Thenthe value of $$\lambda$$ is

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$2$$

Solution

Consider the given vectors,

$$\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k},\overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{c}=\widehat{i}+\widehat{j}-2\overrightarrow{k}$$

Given that projection of $$\overrightarrow{b}+\lambda \overrightarrow{c}$$ on $$\overrightarrow{a}$$ =$$\sqrt{\dfrac{2}{3}}$$

Hence,

$$\dfrac{\left[ \overrightarrow{b}+\lambda \overrightarrow{c} \right].\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$$ =$$\sqrt{\dfrac{2}{3}}$$

$$ \dfrac{\left[ \widehat{i}+2\widehat{j}-\widehat{k}+\lambda \left( \widehat{i}+\widehat{j}-2\widehat{k} \right) \right].\left( 2\widehat{i}-\widehat{j}+\widehat{k} \right)}{\left| 2\widehat{i}-\widehat{j}+\widehat{k} \right|}=\sqrt{\dfrac{2}{3}} $$

$$ \dfrac{\left[ \left( 1+\lambda \right)\widehat{i}+\left( 2+\lambda \right)\widehat{j}+\left( 1-2\lambda \right)\widehat{k} \right].\left( 2\widehat{i}-\widehat{j}+\widehat{k} \right)}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}}}=\sqrt{\dfrac{2}{3}} $$

$$ \dfrac{\left( 1+\lambda \right).2+\left( 2+\lambda \right)\left( -1 \right)+\left( 1-2\lambda \right)\left( 1 \right)}{\sqrt{6}}=\sqrt{\dfrac{2}{3}} $$

$$ 2+2\lambda -2-\lambda +1-2\lambda =\sqrt{\dfrac{2}{3}}.\sqrt{6} $$

$$ -\lambda +1=2 $$

$$ -\lambda =1 $$

$$ \lambda =-1 $$

Hence, this is the answer.
Question 9
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the resultant of two vectors $$\overline {A\,} {\text{and}}\,\overline B $$ makes an angle $$\alpha \,{\text{with}}\,\overline A $$ and $$\beta \,with\,\overline B $$. If $$A = |\overline A |,B = |\overline B |$$, then

A
$$\alpha = \beta $$ always

B
$$\alpha < \beta $$ id A < B

C
$$\alpha < \beta $$ id A > B

D
$$\alpha \ne \beta $$ always

Solution

$$\textbf{Given}:$$ The angle of resultant with Vector A = $$\alpha$$
The angle of vector B with Resultant = $$\beta$$

$$\textbf{Solution}:$$
From the given data, the total angle between Vector A and Vector B = $$\alpha + \beta$$
Now,
$$A = R cos \alpha$$
$$B = R cos\beta$$
If $$ \alpha < \beta$$
Then,
$$cos\alpha > cos\beta$$
$$= R cos\alpha > R cos\beta$$
Hence,
A > B
thus, it can be said
A > B and $$\alpha < \beta$$

$$\textbf{Hence C is the correct option}$$
Question 10
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If $$\overrightarrow a =3\widehat{i}-2\widehat{j}+\widehat{k}$$ and $$\overrightarrow b =2\widehat{i}-4\widehat{j}-3\widehat{k}$$, find $$|\overrightarrow a -2\overrightarrow b|$$.

A
$$86$$

B
$$\sqrt{86}$$

C
$$\sqrt{72}$$

D
$$72$$

Solution

$$\overrightarrow a-2\overrightarrow b=(3\widehat{i}-2\widehat{j}+\widehat{k})-2(2\widehat{i}-4\widehat{j}-3\widehat{k})=-\widehat{i}+6\widehat{j}+7\widehat{k}$$
Therefore,
$$|\overrightarrow a-2\overrightarrow b|=|-\widehat{i}+6\widehat{j}+7\widehat{k}|=\sqrt{(-1)^2+6^2+7^2}=\sqrt{86}$$

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Vector Algebra Test - 50

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Vector Algebra Test - 50

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SHARING IS CARING

Question 1

$$x\epsilon (2, \infty)$$

$$x\epsilon (-\infty, -3)$$

$$x\epsilon (-3, 2)$$

$$x\epsilon \left \{3, 2\right \}$$

Solution

Question 2

$$(Cos \alpha, 2 Sin \alpha)$$

$$(Sin \alpha, Cos \alpha)$$

$$(1, -1)$$

$$(2Cos \alpha, Sin \alpha)$$

Solution

Question 3

$$60^{\circ}$$

$$90^{\circ}$$

$$45^{\circ}$$

$$30^{\circ}$$

Solution

Question 4

$$6\left( b\times c \right)$$

$$\left( a\times b \right)$$

$$6\left( c\times a \right)$$

0

Solution

Question 5

$$2\sqrt 5 $$

$$2\sqrt 2 $$

$$10\sqrt 5 $$

$$5\sqrt 2 $$

Solution

Question 6

$$\large{\frac{\pi}{3}}$$

$$\large{\frac{\pi}{2}}$$

$$\large{\frac{\pi}{4}}$$

$$\large{\frac{2\pi}{3}}$$

Solution

Question 7

$$\vec a \times \left( {\vec b \times \vec c} \right) = 1$$

$$\vec a \times \left( {\vec b \times \vec c} \right) = 0$$

$$\vec a \times \left( {\vec b \times \vec c} \right) = -1$$

None of the above

Solution

Question 8

$$1$$

$$0$$

$$-1$$

$$2$$

Solution

Question 9

$$\alpha = \beta $$ always

$$\alpha < \beta $$ id A < B

$$\alpha < \beta $$ id A > B

$$\alpha \ne \beta $$ always

Solution

Question 10

$$86$$

$$\sqrt{86}$$

$$\sqrt{72}$$

$$72$$

Solution

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