Vector Algebra Test

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Vector Algebra Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Six vectors, a through f have the magnitudes and directions indicated in the figure. Which of the following statements is true?

A
$$b+e=f$$

B
$$b+c=f$$

C
$$d+c=f$$

D
$$d+e=f$$

Solution

Consider the problem
The $$f$$ vector has the resultant of any two other vector.
By parallelogram law of vector addition.
Let $$f$$ be the resultant between $$d$$ and $$e$$
Therefore,
$$d+e=f$$
Question 2
1 / -0

If $$2\vec{a}+3\vec{b}-5\vec{c}=\vec{0}$$, then ratio in which $$\vec{c}$$ divides $$\vec{AB}$$ is

A
$$3:2$$ internally

B
$$3:2$$ externally

C
$$2:3$$ internally

D
$$2:3$$ externally

Solution

Given, $$2\vec{a}+3\vec{b}-5\vec{c}=0$$
$$\Rightarrow \displaystyle\frac{2\vec{a}+3\vec{b}}{5}=\vec{c}$$
$$\Rightarrow \displaystyle\frac{2\vec{a}+3\vec{b}}{2+3}=\vec{c}$$
$$\Rightarrow \displaystyle\frac{\vec{a}+\displaystyle\frac{3}{2}\vec{b}}{1+\displaystyle\frac{3}{2}}=\vec{c}$$ .........$$(i)$$
Let $$\vec{c}$$ divide $$\vec{AB}$$ in the ratio $$\lambda :1$$
Then, $$\vec{c}=\displaystyle\frac{\vec{a}+\lambda \vec{b}}{1+\lambda}$$ ....$$(ii)$$
On comparing Eqs. $$(i)$$ and $$(ii)$$, we get $$\lambda=\displaystyle\frac{3}{2}$$
$$\therefore$$ Required ratio is $$3:2$$ internally.
Question 3
1 / -0

Let $$ABCD$$ be a parallelogram whose diagonals intersect at $$P$$ and $$O$$ be the origin, then $$\vec { OA } +\vec { OB } +\vec { OC } +\vec { OD } $$ equals

A
$$\vec { OP } $$

B
$$2\vec { OP } $$

C
$$3\vec { OP } $$

D
$$4\vec { OP } $$

Solution

Consider the problem
Since
$$P$$ which is the intersection of diagonals of parallelogram its bisects the diagonal
Thus
$$OP=\frac{(OA+OC)}{2}$$
i.e.
$$OA+OC=2\,OP$$ ----- (i)
Similarly
$$OB+OD=OP$$ ----- (ii)

Adding (i) and (ii)
we get
$$\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD} = 4\overrightarrow {OP} $$

Hence the option $$D$$ is the correct answer.
Question 4
1 / -0

The projection of $$\displaystyle \overset { \rightarrow }{ a } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } -2\overset { \wedge }{ k } $$ on $$\displaystyle \overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } $$ is:

A
$$\displaystyle \frac { 1 }{ \sqrt { 14 } } $$

B
$$\displaystyle \frac { 2 }{ \sqrt { 14 } } $$

C
$$\displaystyle \frac{-1}{ \sqrt { 14 } }$$

D
$$\displaystyle \frac { -2 }{ \sqrt { 14 } } $$

Solution

Projection of $$\displaystyle \overset { \rightarrow }{ a } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } -2\overset { \wedge }{ k } $$ on $$\displaystyle \overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } $$ is
$$=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\dfrac{2}{\sqrt {14}}$$
Question 5
1 / -0

If $$\vec {a}$$ is a nonzero vector of magnitude $$'a'$$ and $$\lambda$$ a nonzero scalar, then $$\lambda{\vec {a}}$$ is unit vector if

A
$$\lambda=1$$

B
$$\lambda=-1$$

C
$$a=|\lambda|$$

D
$$a=\dfrac {1}{|\lambda|}$$

Solution

Vector $$\lambda \vec {a}$$ is a unit vector if $$|\lambda \vec {a}|=1$$.
Now,
$$|\lambda \vec {a}|=1$$
$$\Rightarrow |\lambda||\vec {a}|=1$$
$$\Rightarrow |\vec {a}|=\dfrac {1}{|\lambda|}, \; [\lambda \neq 0]$$
$$\Rightarrow a=\dfrac {1}{|\lambda|}, \; [\because |\vec {a}|=a]$$
Hence, vector $$\lambda \vec {a}$$ is a unit vector if $$a=\dfrac {1}{|\lambda|}$$
Thus the correct answer is $$D$$.
Question 6
1 / -0

Find $$u+v$$, when $$u=(3,4,-2)$$ and $$v=(0,-4,0)$$.

A
$$(3,-8,-2)$$

B
$$(-3,0,2)$$

C
$$(3,0,-2)$$

D
None of these

Solution

Given points
$$u(3,4,-2)$$ and $$v(0,-4,0)$$
$$\vec{u}=3\hat{i}+4\hat{j}-2\hat{k}$$
$$\vec{v}=-4\hat{j}$$
vector
$$\vec{u}+\vec{v}=3\hat{i}+4\hat{j}-2\hat{k}-4\hat{j}$$
$$\vec{u}+\vec{v}=3\hat{i}-2\hat{k}=(3,0,-2)$$
Question 7
1 / -0

Find the vector $$w$$ with the initial point $$(9,4)$$ and final point $$(12,6)$$.

A
$$(21,10)$$

B
$$(3,2)$$

C
$$(-21,2)$$

D
none of these

Solution

Given end points
$$A(9,4)$$ and $$B(12,6)$$
$$\vec{OA}=9\hat{i}+4\hat{j}$$
$$\vec{OB}=12\hat{i}+6\hat{j}$$
vector w from A to B
$$\vec{AB}=\vec{OB}-\vec{OA}$$
$$\vec{w}=12\hat{i}+6\hat{j}-(9\hat{i}+4\hat{j})$$
$$\vec{w}=12\hat{i}+6\hat{j}-9\hat{i}-4\hat{j}$$
$$\vec{w}=3\hat{i}+2\hat{j}$$
$$=(3,2)$$
Question 8
1 / -0

Give the vector from $$(2,-7,0)$$ to $$(1,-3,-5)$$.

A
$$(3,4,-5)$$

B
$$(-1,4,-5)$$

C
$$(3,-10,-5)$$

D
None of these

Solution

Given points
$$A(2,-7,0)$$ and $$B(1,-3,-5)$$
$$\vec{OA}=2\hat{i}-7\hat{j}$$
$$\vec{OB}=\hat{i}-3\hat{j}-5\hat{k}$$
vector from A to B
$$\vec{AB}=\vec{OB}-\vec{OA}$$
$$\vec{AB}=\hat{i}-3\hat{j}-5\hat{k}-(2\hat{i}-7\hat{j})$$
$$\vec{AB}=\hat{i}-3\hat{j}-5\hat{k}-2\hat{i}+7\hat{j}$$
$$\vec{AB}=-\hat{i}+4\hat{j}-5\hat{k}$$
$$AB=(-1,4,-5)$$
Question 9
1 / -0

Find the vector which joins the point A$$(4,5,6)$$ to B$$(10,11,12)$$.

A
$$14\hat i+16\hat j+18\hat k$$

B
$$-6\hat i-6\hat j-6\hat k$$

C
$$6\hat i+6\hat j+6\hat k$$

D
None of these

Solution

Given points
$$A(4,5,6)$$ and $$B(10,11,12)$$
$$\vec{OA}=4\hat{i}+5\hat{j}+6\hat{k}$$
$$\vec{OB}=10\hat{i}+11\hat{j}+12\hat{k}$$
vector
$$\vec{AB}=\vec{OB}-\vec{OA}$$
$$\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-(4\hat{i}+5\hat{j}+6\hat{k})$$
$$\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-4\hat{i}-5\hat{j}-6\hat{k}$$
$$\vec{AB}=6\hat{i}+6\hat{j}+6\hat{k}$$
Question 10
1 / -0

Give the vector from $$(1,-3,-5)$$ to $$(2,-7,0)$$.

A
$$(1,-4,5)$$

B
$$(3,-10,-5)$$

C
$$(1,4,5)$$

D
None of these

Solution

Given points
$$A(1,-3,-5)$$ and $$B(2,-7,-0)$$
$$\vec{OA}=\hat{i}-3\hat{j}-5\hat{k}$$
$$\vec{OB}=2\hat{i}-7\hat{j}$$
vector from $$A$$ to $$B$$
$$\vec{AB}=\vec{OB}-\vec{OA}$$
$$\vec{AB}=2\hat{i}-7\hat{j}-(\hat{i}-3\hat{j}-5\hat{k})$$
$$\vec{AB}=2\hat{i}-7\hat{j}-\hat{i}+3\hat{j}+5\hat{k}$$
$$\vec{AB}=\hat{i}-4\hat{j}+5\hat{k}$$
$$AB=(1,-4,5)$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 17

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Vector Algebra Test - 17

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SHARING IS CARING

Question 1

$$b+e=f$$

$$b+c=f$$

$$d+c=f$$

$$d+e=f$$

Solution

Question 2

$$3:2$$ internally

$$3:2$$ externally

$$2:3$$ internally

$$2:3$$ externally

Solution

Question 3

$$\vec { OP } $$

$$2\vec { OP } $$

$$3\vec { OP } $$

$$4\vec { OP } $$

Solution

Question 4

$$\displaystyle \frac { 1 }{ \sqrt { 14 } } $$

$$\displaystyle \frac { 2 }{ \sqrt { 14 } } $$

$$\displaystyle \frac{-1}{ \sqrt { 14 } }$$

$$\displaystyle \frac { -2 }{ \sqrt { 14 } } $$

Solution

Question 5

$$\lambda=1$$

$$\lambda=-1$$

$$a=|\lambda|$$

$$a=\dfrac {1}{|\lambda|}$$

Solution

Question 6

$$(3,-8,-2)$$

$$(-3,0,2)$$

$$(3,0,-2)$$

None of these

Solution

Question 7

$$(21,10)$$

$$(3,2)$$

$$(-21,2)$$

none of these

Solution

Question 8

$$(3,4,-5)$$

$$(-1,4,-5)$$

$$(3,-10,-5)$$

None of these

Solution

Question 9

$$14\hat i+16\hat j+18\hat k$$

$$-6\hat i-6\hat j-6\hat k$$

$$6\hat i+6\hat j+6\hat k$$

None of these

Solution

Question 10

$$(1,-4,5)$$

$$(3,-10,-5)$$

$$(1,4,5)$$

None of these

Solution

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