Vector Algebra Test

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Vector Algebra Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

The vector $$b = 3j + 4k$$ is to be written as the sum of a vector $$b_{1}$$ parallel to $$a = i + j$$ and a vector $$b_{2}$$ perpendicular to $$a$$. Then $$b_{1}$$ is equal to

A
$$\dfrac {3}{2}(i + j)$$

B
$$\dfrac {2}{3}(i + j)$$

C
$$\dfrac {1}{2}(i + j)$$

D
$$\dfrac {1}{3}(i + j)$$

Solution

$$b_{1} \parallel a\Rightarrow b_{1} = a(i + j)$$
$$b_{2} = b - b_{1} = (3 - a) i - aj + 4k$$
Also, $$b_{3} - a = 0$$
$$\Rightarrow (3 - a) - a=0\Rightarrow a = \dfrac {3}{2}$$
$$\therefore b_{1} = \dfrac {3}{2}(i + j)$$
Question 2
1 / -0

The magnitude of the scalar $$p$$ for which the vector $$p\left( -3\hat { i } -2\hat { j } +13\hat { k } \right) $$ is of unit length is:

A
$$\dfrac{1}{8}$$

B
$$\dfrac{1}{64}$$

C
$$\sqrt { 182 } $$

D
$$\dfrac{1}{\sqrt { 182 }} $$

Solution

For given vector to be unit vector its magnitude must be equal to one.
Now modulus of given vector $$=p\sqrt { { (-3) }^{ 2 }+{ (-2) }^{ 2 }+{ 13 }^{ 2 } } =p\sqrt { 182 } =1\\ \Rightarrow p=\dfrac { 1 }{ \sqrt { 182 } } $$
Option D is correct.
Question 3
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If $$\vec a, \vec b$$ and $$\vec c$$ are three non-coplanar vectors, then $$(\vec a +\vec b -\vec c) \cdot [(\vec a - \vec b) \times (\vec b - \vec c)$$ equals

A
$$\vec 0$$

B
$$\vec a.(\vec b\times \vec c)$$

C
$$\vec a.(\vec c\times \vec b)$$

D
$$3\vec a.(\vec b\times \vec c)$$

Solution

$$[a + b - c] \cdot [(a - b)\times (b - c)$$

$$= (a + b - c)\cdot [a\times b - a\times c - b\times b + b\times c]$$

Since, $$(b \times b=0)$$

$$\therefore [a + b - c] \cdot [(a - b)\times (b - c)]$$
$$= a.(a\times b) - a.(a\times c) + a.(b\times c) + b.(a\times b) - b(a\times c) + b.(b\times c) - c.(a\times b) + c.(a\times c) - c.(b\times c)$$

$$= a.(b\times c) - b.(a\times c) - c.(a\times b)$$ ........ (since, a,b and c are non-
coplanar)
$$=[a\ b\ c] - [b\ a\ c] - [c\ a\ b]$$
$$=[a\ b\ c] + [a\ b\ c] - [a\ b\ c]$$

$$=[a\ b\ c] = a.(b\times c)$$

Hence, option B is correct.
Question 4
1 / -0

Let $$\vec{v_1}, \vec{v_2}, \vec{v_3} , \vec{v_4} $$ be unit vectors in the xy - plane, one each in the interior of the four quadrants. Which of the following statements is necesserily true?

A
$$\vec{v_1}, \vec{v_2}, \vec{v_3} , \vec{v_4} = 0$$

B
There exist i, j with $$ 1 \le i \le j \le 4 $$ such that $$ \vec{v_i} + \vec{v_j} $$ is in the first quadrant.

C
There exist i, j with $$ 1 \le i \le j \le 4 $$ such that $$ \vec{v_i} + \vec{v_j} < 0$$

D
There exist i, j with $$ 1 \le i \le j \le 4 $$ such that $$ \vec{v_i} + \vec{v_j} >0 $$

Solution

A zero vector can never be a unit vector

$$\therefore $$ option $$A$$ is not correct

Let as assume four unit vectors in the $$xy$$ plane as shown in the figure

$$\underset { { V }_{ 1 } }{ \rightarrow } =\dfrac { i }{ \sqrt { 2 } } +\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 2 } }{ \rightarrow } =-\dfrac { i }{ \sqrt { 2 } } +\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 3 } }{ \rightarrow } =-\dfrac { i }{ \sqrt { 2 } } -\dfrac { j }{ \sqrt { 2 } } \\ \underset { { V }_{ 4 } }{ \rightarrow } =\dfrac { i }{ \sqrt { 2 } } -\dfrac { j }{ \sqrt { 2 } } $$

If we add up any of the two vectors it will either lies on the $$x$$ aisx or the $$y$$ axis, none of them lies in the first quardrant

$$\therefore $$ option $$B$$ is not correct

Sum of any two vectors less than or greater than zero does not make any sense

$$\therefore $$ options $$C$$ and $$D$$ are incorrect

So none of the options are correct.
Question 5
1 / -0

If the position vector $$\overrightarrow{a}$$ of the point $$(5, n)$$ is such that $$|\overrightarrow{a}|=13$$, then the value/values of n be

A
$$\pm 8$$

B
$$\pm 12$$

C
8 only

D
12 only

Solution

Solution:
Given that: $$|\overrightarrow a|=13$$
We have,
$$\overrightarrow a=5\hat i+n\hat j$$
or, $$|\overrightarrow a|=\sqrt{5^2+n^2}$$
or, $$13=\sqrt{25+n^2}$$
or, $$n^2=169-25=144$$
or, $$n=\pm12$$
Hence, B is the correct option.
Question 6
1 / -0

If $$\vec { a } ,\vec { b } ,\vec { c } $$ are mutually perpendicular unit vectors, then $$\left| \vec { a } +\vec { b } +\vec { c } \right| $$ is equal to

A
$$3$$

B
$$\sqrt { 3 } $$

C
Zero

D
$$1$$

Solution

Solution:

Given that: $$\vec a,\vec b,\vec c$$ are mutually perpendicular .i.e
$$\vec a.\vec b=0,\vec b.\vec c=0$$ and $$\vec c.\vec a=0$$
Also, $$|\vec a|=|\vec b|=|\vec c|=1$$

Solution:
$$|\vec a+\vec b+\vec c|=\sqrt{|\vec a|^2+|\vec b|^2+|\vec c|^2+2\vec a.\vec b+2\vec b.\vec c+2\vec c.\vec a}$$
$$=\sqrt{1+1+1+0+0+0}$$

$$=\sqrt{3}$$

Hence, B is the correct option.
Question 7
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Let $$u, v$$ and $$w$$ be such that $$\left| u \right| =1, \left| v \right| =3$$ and $$\left| w \right| =2$$. If the projection of $$v$$ along $$u$$ is equal to that of $$w$$ along $$u$$ and vectors $$v$$ and $$w$$ are perpendicular to each other, then $$\left| u-v+w \right| $$ equals

A
$$2$$

B
$$\sqrt { 7 } $$

C
$$\sqrt { 14 } $$

D
$$14$$

Solution

Given, $$v\cdot u=w\cdot u$$ and $$v \perp w$$ $$\Rightarrow v\cdot w=0$$
Now, consider
$${ \left| u-v+w \right| }^{ 2 }={ \left| u \right| }^{ 2 }+{ \left| v \right| }^{ 2 }+{ \left| w \right| }^{ 2 }-2u\cdot v-2w\cdot v+2u\cdot w$$
$$=1+9+4=14$$
$$\Rightarrow \left| u-v+w \right| =\sqrt { 14 } $$
Question 8
1 / -0

A vector $$R$$ is given by $$R=A\times \left( B\times C \right) $$, which of the following is true?

A
$$R$$ must be perpendicular to $$B$$

B
$$R$$ is parallel to $$A$$

C
$$R$$ must be parallel to $$B$$

D
None of the above

Solution

Given that:
$$R=A\times(B\times C)$$
Geometrical representation of the above is
$$R$$ must be perpendicular to $$A$$ and also $$R$$ must be perpendicular to $$(B\times C).$$
Therefore, D is the correct option.
Question 9
1 / -0

If $$a=\hat{i}+\hat{j}, b=2\hat{j}-\hat{k}$$ and $$r\times a=b\times a, r\times b=a\times b$$, then a unit vector in the direction of $${r}$$ is?

A
$$\displaystyle\frac{1}{\sqrt{11}}(\hat{i}+3\hat{j}-\hat{k})$$

B
$$\displaystyle \frac{1}{\sqrt{11}}(\hat{i}-3\hat{j}+\hat{k})$$

C
$$\displaystyle \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$$

D
None of these

Solution

We have, $$r\times a =b\times a$$ and $$r\times b=a\times b$$

$$\Rightarrow r\times a =-(r\times b)$$
$$\Rightarrow r\times (a+b)=0$$

$$\Rightarrow $$ r is parallel to $$a+b$$
$$\Rightarrow r=\lambda(a+b)$$

$$\Rightarrow r=\lambda(\hat{i}+3\hat{j}-\hat{k})$$

$$\Rightarrow |r|=\sqrt{11}\lambda$$

$$\therefore$$ Required vector $$=\displaystyle\frac{r}{|r|}=\frac{1}{\sqrt{11}}(\hat{i}+3\hat{j}-\hat{k})$$.
Question 10
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The resultant of $$P$$ and $$Q$$ is $$R$$. If $$Q$$ is doubled, $$R$$ is also doubled and if $$Q$$ is reversed, $$R$$ is again doubled. Then, $$P^{2} : Q^{2} : R^{2}$$ given by

A
$$2 : 2 : 3$$

B
$$3 : 2 : 2$$

C
$$2 : 3 : 2$$

D
$$2 : 3 : 1$$

Solution

$$R^{2} = P^{2} + Q^{2} = 2PQ\cos \theta$$
$$4R^{2} = P^{2} + 4Q^{2} + 4PQ \cos \theta$$
$$4R^{2} = P^{2} + 4Q^{2} - 2PQ \cos \theta$$
On solving,
$$\dfrac {P^{2}}{-6} = \dfrac {Q^{2}}{-9} = \dfrac {R^{2}}{-6}$$
$$\Rightarrow P^{2} : Q^{2} : R^{2}$$ as $$2 : 3 : 2$$.

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 45

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Vector Algebra Test - 45

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SHARING IS CARING

Question 1

$$\dfrac {3}{2}(i + j)$$

$$\dfrac {2}{3}(i + j)$$

$$\dfrac {1}{2}(i + j)$$

$$\dfrac {1}{3}(i + j)$$

Solution

Question 2

$$\dfrac{1}{8}$$

$$\dfrac{1}{64}$$

$$\sqrt { 182 } $$

$$\dfrac{1}{\sqrt { 182 }} $$

Solution

Question 3

$$\vec 0$$

$$\vec a.(\vec b\times \vec c)$$

$$\vec a.(\vec c\times \vec b)$$

$$3\vec a.(\vec b\times \vec c)$$

Solution

Question 4

$$\vec{v_1}, \vec{v_2}, \vec{v_3} , \vec{v_4} = 0$$

There exist i, j with $$ 1 \le i \le j \le 4 $$ such that $$ \vec{v_i} + \vec{v_j} $$ is in the first quadrant.

There exist i, j with $$ 1 \le i \le j \le 4 $$ such that $$ \vec{v_i} + \vec{v_j} < 0$$

There exist i, j with $$ 1 \le i \le j \le 4 $$ such that $$ \vec{v_i} + \vec{v_j} >0 $$

Solution

Question 5

$$\pm 8$$

$$\pm 12$$

8 only

12 only

Solution

Question 6

$$3$$

$$\sqrt { 3 } $$

Zero

$$1$$

Solution

Question 7

$$2$$

$$\sqrt { 7 } $$

$$\sqrt { 14 } $$

$$14$$

Solution

Question 8

$$R$$ must be perpendicular to $$B$$

$$R$$ is parallel to $$A$$

$$R$$ must be parallel to $$B$$

None of the above

Solution

Question 9

$$\displaystyle\frac{1}{\sqrt{11}}(\hat{i}+3\hat{j}-\hat{k})$$

$$\displaystyle \frac{1}{\sqrt{11}}(\hat{i}-3\hat{j}+\hat{k})$$

$$\displaystyle \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$$

None of these

Solution

Question 10

$$2 : 2 : 3$$

$$3 : 2 : 2$$

$$2 : 3 : 2$$

$$2 : 3 : 1$$

Solution

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