Vector Algebra Test

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Vector Algebra Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The ratio in which the line joining $$(2,-4,3)$$ and $$(-4,5,-6)$$ is divided by the plane $$3x+2y+z-4=0$$ is

A
$$2 : 1$$

B
$$4 : 3$$

C
$$1 : 4$$

D
$$ 2 : 3$$

Solution

We have given points as shown in the figure
$$Eq{u^n}\,of\,line\,\,AB:\,\frac{{x - 2}}{6} = \frac{{y + 4}}{{ - 9}} = \frac{{z - 3}}{9}$$
Now
Points $$P$$ is $$\left( {6r + 2,\, - 9r - 4,\,9r + 3} \right)$$
$$3\left( {6r + 2} \right) + 2\left( { - 9r - 4} \right) + 9r + 3 = 4$$
$$9r + 1 = 4$$
$$\therefore r = \dfrac{1}{3}$$
So,
$$P\left( {4, - 7,6} \right)$$
Now,
$$\frac{{AP}}{{BP}} = \sqrt {\frac{{{{\left( { - 2} \right)}^2} + {{\left( 3 \right)}^2} + {{\left( { - 3} \right)}^2}}}{{{{\left( 8 \right)}^2} + {{\left( {12} \right)}^2} + {{\left( 12 \right)}^2}}}} $$
$$ = \sqrt {\frac{{22}}{{352}}} $$
$$ = \sqrt {\frac{{11}}{{176}}} $$
$$ = \sqrt {\frac{1}{{16}}} = \frac{1}{4}$$
As $$P$$ divides externally in the ratio of $$1:4$$.
Hence, The option $$C$$ is the correct answer.
Question 2
1 / -0

A vector $$\overrightarrow { A } $$ points vertically downward(south)and $$\overrightarrow { B } $$ points towards east, then the vector product $$\overrightarrow { A } \times \overrightarrow { B } $$ is:

A
Along west

B
Along east

C
Zero

D
Outwards or inwards

Solution

$$\vec{A}\times \vec{B}$$ is perpendicular to the plane of paper and acting outward. It is denoted by $$\otimes $$
Question 3
1 / -0

$$\bar{a}$$ and $$\bar{b}$$ are the position vectors of A and B respectively. Points P and Q divide AB, internally and externally in the same ratio 2:1 , then PQ is equal to

A
$$\frac{3}{2}(\bar{b}-\bar{a})$$

B
$$\frac{4}{3}(\bar{a}-\bar{b})$$

C
$$\frac{5}{6}(\bar{b}-\bar{a})$$

D
$$\frac{4}{3}(\bar{b}-\bar{a})$$

Solution

Position vector of P will be $$\vec{OP}=\left(\dfrac{2\bar{b}+\bar{a}}{3}\right)$$ (P divides AB internally in the ratio $$2:1$$)
Position vector of Q will be $$\vec{OQ}=\left(\dfrac{2\bar{b}-\bar{a}}{2-1}\right)=\left(\dfrac{2\bar{b}-\bar{a}}{1}\right)$$
(Q divides AB externally in the ratio $$2:1$$)
$$\vec{PQ}=\vec{OQ}-\vec{OP}=\left(2\vec{b}-\vec{a}\right)-\left(\dfrac{2\bar{b}+\bar{a}}{3}\right)=\dfrac{4\vec{b}}{3}-\dfrac{4\vec{a}}{3}$$
$$=\dfrac{4}{3}(\vec{b}-\vec{a})$$.
Question 4
1 / -0

Two vector $$A$$ and $$B$$ have equal magnitudes. Then the vector $$A+B$$ is perpendicular to

A
$$A\times{B}$$

B
$$A-B$$

C
$$3A-3B$$

D
All of these

Solution

$$\vec{A}\times \vec{B}$$ is a vector perpendicular to plane $$\vec{A}+\vec{B}$$ and hence perpendicular to $$\vec{A}+\vec{B}$$
Question 5
1 / -0

The work done by the force $$\vec { F } = 2 \hat { i } - \hat { j } - \hat { \mathbf { k } }$$ in moung an object along the vector $$3 \hat { i } + 2 j - 5 \hat { k }$$ is

A
$$- 9$$ units

B
$$9$$ units

C
$$-15$$ units

D
None of these

Solution

$$\vec { F } = 2 \hat { i } - \hat { j } - \hat { k } $$
$$\vec { r } = 3 \hat { i } +2 \hat { j } - 5\hat { k } $$
$$Work=\vec { F }.\vec{r}$$
$$Work= (2\hat { i } - \hat { j } - \hat { k }).(3 \hat { i } +2 \hat { j } - 5\hat { k })$$
$$Work=6-2+5$$
$$Work=11-2=9units$$
Hence the correct option is (B).
Question 6
1 / -0

If $$2\overline a - 4\widehat i - 2\widehat j + \widehat k = 0$$ then find $$\overline a $$.

A
$$2 \overline { j } + \overline { \mathbf { k } }$$

B
$${ 2\widehat { i } +\widehat { j } -\frac { 1 }{ 2 } \widehat { k } }$$

C
$$2 \overline { i } + \overline { \mathbf { j } }$$

D
$$2$$ $$\overline { i - j }$$

Solution

$$\begin{array}{l} 2\overline { a } -4\widehat { i } -2\widehat { j } +\widehat { k } =0 \\ 2\overline { a } =4\widehat { i } +2\widehat { j } -\widehat { k } \\ \overline { a } =\left( { 2\widehat { i } +\widehat { j } -\frac { 1 }{ 2 } \widehat { k } } \right) \end{array}$$
Question 7
1 / -0

Let $$\vec{a}=\hat{i}+\hat{j}+\hat{k}$$ and $$\vec{b}$$ is a vector such that $$\vec{a}.\vec{b}=0$$ and $$\vec{a}\times \vec{b}=0$$. Then which of following is correct?

A
$$\vec{b}=0$$

B
$$\vec{b} \perp\vec{a}$$

C
$$\vec{b}$$ is non-zero vector

D
$$\vec{b}$$ is parallel to $$\vec{a}$$

Solution

$$\begin{array}{l} \vec { a } .\vec { b } =0 \\ \vec { a } \times \vec { b } =0 \\ \left| { \vec { a } } \right| \left| { \vec { b } } \right| \cos \theta =0 \\ \left| { \vec { b } } \right| \cos \theta =0 \\ or, \\ \vec { a } =\hat { i } +\hat { j } +\hat { k } \\ \left| { \vec { a } } \right| =\sqrt { 3 } \\ \left| { \vec { a } } \right| \left| { \vec { b } } \right| \sin \theta =0 \\ \left| { \vec { b } } \right| \sin \theta =0 \\ N0w, \\ \left| { \vec { b } } \right| =0 \\ \vec { b } =0 \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$$
Question 8
1 / -0

Given that $$\vec{ A } \times \vec{ B } =\vec{ B } \times \vec { C } =\vec { 0 } $$ if $$\vec{ A } \vec { B } \vec { C } $$ are not null vectors, Find the value of $$\vec{ A } \times \vec{ C } $$

A
$$\vec { A } \times \vec { B } $$

B
$$\vec { 0 } $$

C
$$\vec{ C } \times \vec { B } $$

D
$$\vec { C } \times \vec { A } $$

Solution

We have,
$$\begin{array}{l} \vec { A } \times \vec { B } =\vec { B } \times \vec { C } =\vec { 0 } \\ \vec { A } \times \vec { B } +\vec { C } \times \vec { B } =\vec { 0 } \\ \left( { \vec { A } +\vec { C } } \right) \times \vec { B } =\vec { 0 } \\ \vec { A } +\vec { C } =\lambda \vec { B } \\ \vec { A } \times \vec { C } +0=\lambda \left( { \vec { B } \times \vec { C } } \right) \\ \vec { A } \times \vec { C } =\vec { 0 } \\ Hence,\, the\, option\, B\, is\, correct.\, \end{array}$$
Question 9
1 / -0

Magnitude of vector $$3\hat { i } -12\hat { j } -4\hat { k } $$

A
13

B
$$\sqrt { 13 } $$

C
19

D
none

Solution

Let $$ a=3\hat{i}-12\hat{j}-4\hat{k}$$
Magnitude of a,$$\left | a \right |$$ =$$\sqrt{3^{2}+(-12)^{2}+(-4)^{2}}$$
$$ = \sqrt{9+144+16}$$
$$ = \sqrt{169}$$
$$ =13.$$
Question 10
1 / -0

$$\vec{r} = \vec{x}\hat{i}+\vec{y}\hat{j}$$ is the equation of:

A
$$yoz$$ plane

B
a straight line joining the points $$\vec{i}$$ and $$\vec{j}$$

C
$$zox$$ plane

D
$$xoy$$ plane

Solution

A vector is defined by its magnitude and its orientation with respect to a set of coordinates. It is often useful in analyzing vectors to break them into their component parts. For two-dimensional vectors, these components are horizontal and vertical.
Vector can always be represented in two dimension and three dimension. In two dimension, vector is represented in the form of $$x\hat{i}+y\hat{j}$$. In three dimension, it is represented by $$x\hat{i} + y\hat{j} + z\hat{k}$$.
Thus, $$\overrightarrow{r}=x\hat{i}+y\hat{j}$$ is vector in $$xy$$ plane.

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 22

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Vector Algebra Test - 22

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SHARING IS CARING

Question 1

$$2 : 1$$

$$4 : 3$$

$$1 : 4$$

$$ 2 : 3$$

Solution

Question 2

Along west

Along east

Zero

Outwards or inwards

Solution

Question 3

$$\frac{3}{2}(\bar{b}-\bar{a})$$

$$\frac{4}{3}(\bar{a}-\bar{b})$$

$$\frac{5}{6}(\bar{b}-\bar{a})$$

$$\frac{4}{3}(\bar{b}-\bar{a})$$

Solution

Question 4

$$A\times{B}$$

$$A-B$$

$$3A-3B$$

All of these

Solution

Question 5

$$- 9$$ units

$$9$$ units

$$-15$$ units

None of these

Solution

Question 6

$$2 \overline { j } + \overline { \mathbf { k } }$$

$${ 2\widehat { i } +\widehat { j } -\frac { 1 }{ 2 } \widehat { k } }$$

$$2 \overline { i } + \overline { \mathbf { j } }$$

$$2$$ $$\overline { i - j }$$

Solution

Question 7

$$\vec{b}=0$$

$$\vec{b} \perp\vec{a}$$

$$\vec{b}$$ is non-zero vector

$$\vec{b}$$ is parallel to $$\vec{a}$$

Solution

Question 8

$$\vec { A } \times \vec { B } $$

$$\vec { 0 } $$

$$\vec{ C } \times \vec { B } $$

$$\vec { C } \times \vec { A } $$

Solution

Question 9

13

$$\sqrt { 13 } $$

19

none

Solution

Question 10

$$yoz$$ plane

a straight line joining the points $$\vec{i}$$ and $$\vec{j}$$

$$zox$$ plane

$$xoy$$ plane

Solution

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