Vector Algebra Test

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Vector Algebra Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

$$ABCD$$ is a quadrilateral and $$E$$ the point of intersection of the lines joining the middle points of opposite sides. If $$O$$ is any point, then the resultant of $$OA,OB,OC$$ and $$OD$$ is equal to

A
$$2OE$$

B
$$OE$$

C
$$4OE$$

D
none of these

Solution

Let $$P,Q,R,S$$ be the mid-points of sides $$BC,CD,DA,AB$$ respectively of a quadrilateral $$ABCD$$.
BY geometry the figure formed by joining the mid-point $$P,Q,R,S$$ will be a parallelogram.
Hence, its diagonal will bisect each other, say at $$E$$.
Now $$\because P$$ is the mid point of $$BC$$
$$\therefore OB+OC=2OP$$ ....(1)
And $$R$$ is the mid point of $$AD$$
$$\therefore OA+OD=2OR$$ ....(2)
Adding (1) and (2), we have
$$OA+OB+OC+OD=2(OP+OR)=2.2OE=4OE$$
$$(\because E$$ is mid point of $$PR$$,
$$\therefore OP+OR=2OE)$$.
Question 2
1 / -0

The components of vector $$ i + j + k $$ along vector $$i+2j+3k $$ is

A
$$(3/7) (i + 2j + 3k)$$

B
$$(i + 2j + 3k)$$

C
$$(1/7) (i + 2j + 3k)$$

D
$$(4/7) (i + 2j + 3k)$$

Solution

The projection of vector $$ i + j + k $$ along vector, $$i+2j+3k $$is
$$\quad =\displaystyle \frac{(i+j+k)\cdot (i+2j+3k)}{|i+2j+3k|}=\frac{6}{\sqrt{14}}$$
Hence component of vector $$i+j+k$$ along $$i+2j+3k$$ is,
$$\quad =\displaystyle \frac{6}{\sqrt{14}}.\frac{i+2j+3k}{|1+2j+3k|}=\frac{3}{7}(i+2j+3k)$$
Question 3
1 / -0

OAB is a given triangle such that $$\displaystyle \overrightarrow{OA}=\bar a$$, $$\displaystyle \overrightarrow{OB}=\bar b$$.
Also C is a point on AB such that $$\overrightarrow{AB}=2\overrightarrow{BC}$$. State which of the following statements are correct?

A
$$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left ( \bar b-\bar a \right )$$,

B
$$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left ( \bar a-\bar b \right )$$,

C
$$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left (\bar b+\bar a \right )$$,

D
$$\displaystyle \overrightarrow{AC}=\frac{3}{2}\left ( \bar b-\bar a \right )$$,

Solution

Given $$\overline{OA}=\bar a$$ and $$\overline{OB}=\bar b$$
$$C$$ is apoint on $$AB$$ such that $$\overline{AB}=2\overline{BC}$$
$$\Rightarrow \bar b - \bar a = 2(\bar c - \bar b)$$
$$\Rightarrow \bar c=\displaystyle\dfrac{3\bar b-\bar a}{2}$$
$$\therefore \overline{AC}=\bar c-\bar a =\displaystyle\dfrac{3}{2}(\bar b-\bar a)$$
Hence, option D.
Question 4
1 / -0

Let $$\displaystyle \overrightarrow{OA}=a$$ and $$\displaystyle \overrightarrow{OB}=b$$ and $$\displaystyle \overrightarrow{OC}=a+b$$. What is the type of the quadrilateral $$OACB$$?

A
Parallelogram

B
Trapezium

C
Rectangle

D
Both A and C

Solution

Given that $$\vec{OA}=a$$, $$\vec{OB}=b$$ and $$\vec{OC}=a+b$$
and $$\vec{BC}=\vec{OC}-\vec{OB}=a$$
and $$\vec{CA}=\vec{OA}-\vec{OC}=-b$$
Here, $$|\vec{OA}|=|\vec{BC}|$$ and $$OA$$ is parallel to $$BC$$
Similarly for $$\vec{OB}$$ and $$\vec{CA}$$
Therefore, $$OABC$$ is both parallelogram and rectangle.
Question 5
1 / -0

A vector $$a$$ can be written as

A
$$(a\cdot i)i+(a\cdot j)j+(a\cdot k)k$$

B
$$(a\cdot j)i+(a\cdot k)j+(a\cdot i)k$$

C
$$(a\cdot k)j+(a\cdot i)j+(a\cdot j)k$$

D
$$(a\cdot a)i+(i+j+k) k$$

Solution

Let $$a=a_{1} i+a_{2} i+a_{3} k$$, so that $$a.i=a
_{1},a\cdot j=a_{2}$$ and $$ a.k=a_{3}$$
$$\Rightarrow a=(a\cdot i)i+(a\cdot j)j+(a\cdot k)k$$
Question 6
1 / -0

The projection of $$\bar a=3\hat i-\hat j +5\hat k$$ on $$\bar b=2\hat i+3\hat j+\hat k$$ is

A
$$\displaystyle 8/\sqrt{(35)}$$

B
$$\displaystyle 8/\sqrt{(39)}$$

C
$$\displaystyle 8/\sqrt{(14)}$$

D
$$\displaystyle \sqrt{(14)}$$

Solution

The projection of $$\bar a=3\hat i-\hat j +5\hat k$$ on $$\bar b=2\hat i+3\hat j+\hat k$$ is $$\displaystyle\frac{\bar a\cdot \bar b}{|\bar b|}$$
$$=\displaystyle\frac{(3\hat i-\hat j +5\hat k)\cdot (2\hat i+3\hat j+\hat k) }{|2\hat i+3\hat j+\hat k|}$$
$$=\displaystyle\frac{8}{\sqrt {14}}$$
Hence, option C.
Question 7
1 / -0

The triangle $$ABC$$ is defined by the vertices $$A (1, -2, 2), B(1, 4, 0)$$ and $$C(-4, 1, 1)$$. Let $$M$$ be the foot of the altitude drawn from the vertices $$B$$ to side $$AC$$. Then $$\displaystyle \overrightarrow{BM}=$$

A
$$\left (-\dfrac {20}{7}, -\dfrac {30}{7}, \dfrac {10}{7}\right)$$

B
$$(-20, -30, 10)$$

C
$$(2, 3, -1)$$

D
none of these

Solution

$$\displaystyle\overrightarrow { MB } =\overrightarrow { AB } -\overrightarrow { AM } =\overrightarrow { AB } -\dfrac { \left( \overrightarrow { AB } .\overrightarrow { AC } \right) \overrightarrow { AC } }{ { \left( \overrightarrow {| AC |} \right) }^{ 2 } }$$
$$\displaystyle=\left( 6\vec j-2\vec k \right) -\dfrac { \left( 6\vec j-2\vec k \right) .\left( 5\vec i+3\vec j-\vec k \right) \left( -5\vec i+3\vec j-\vec k \right) }{ \left( 25+9+1 \right) }$$
$$\displaystyle=\left( 6\vec j-2\vec k \right) -\dfrac { 20 }{ 35 } \left( -5\vec i+3\vec j-\vec k \right)$$ $$\displaystyle=\dfrac { 10 }{ 7 } \left( 2\vec i+3\vec j-k \right)$$
$$\displaystyle\therefore \overrightarrow { BM } =-\dfrac { 10 }{ 7 } \left( 2\vec i+3\vec j-\vec k \right) =\left( -\dfrac { 20 }{ 7 } \vec i-\dfrac { 30 }{ 7 } \vec j+\dfrac { 10 }{ 7 }\vec k \right) $$
Question 8
1 / -0

Given two vectors $$a=2\hat{i}-3\hat{j}+6\hat{k}$$, $$b=-2\hat{i}+2\hat{j}-1\hat{k}$$ and $$\displaystyle \lambda$$ = ratio of the projection of $$a$$ on $$b$$ and the projection of $$b$$ on $$a,$$ then the value of $$\displaystyle \lambda $$ is

A
$$\dfrac{3}{7}$$

B
$$\dfrac{7}{3}$$

C
$$3$$

D
$$7$$

Solution

Projection of $$\vec{b}$$ on $$\vec{a}$$ will be
$$=bcos\theta$$
$$=\dfrac{\vec{a}.\vec{b}}{a}$$ ...(i)
Similarly
Projection of $$\vec{a}$$ on $$\vec{b}$$ will be
$$=acos\theta$$
$$=\dfrac{\vec{a}.\vec{b}}{b}$$ ...(ii)
Therefore
$$\lambda=\dfrac{ii}{i}$$

$$=\dfrac{\dfrac{\vec{a}.\vec{b}}{b}}{\dfrac{\vec{a}.\vec{b}}{a}}$$

$$=\dfrac{a}{b}$$

$$=\dfrac{\sqrt{2^{2}+3^{2}+6^{2}}}{\sqrt{2^{2}+2^{2}+1^{2}}}$$

$$=\dfrac{\sqrt{49}}{\sqrt{9}}$$

$$=\dfrac{7}{3}$$
Question 9
1 / -0

Let $$a=i+2j+k, \ b=i-j+k$$ and $$c=i+j-k $$. A vector in the plane of $$ a$$ and $$b$$, whose projection on $$c$$ is $$\displaystyle \frac{1}{\sqrt{3}}$$ is

A
$$-4i+j-4k$$

B
$$3i+j-3k$$

C
$$ i+j-2k$$

D
$$4i+j-4k$$

Solution

A vector $$d$$ in the plane of $$a$$ and $$b$$ is given by, $$d =\lambda

a+\mu b =(\lambda +\mu)i+(2\lambda -\mu)j+(\lambda+\mu)k,$$
Now the projection of $$d$$on $$c$$ is $$\displaystyle \frac{c\cdot

d}{\left|c\right|}=\frac{\lambda

+\mu+(2\lambda-\mu)-(\lambda+\mu)}{\sqrt{3}}=\frac{1}{\sqrt{3}}$$(given)
$$\Rightarrow 2\lambda

-\mu=1$$. Thus $$d=(3\lambda-1)i+j+(3\lambda-1)k, $$
So for $$ \lambda =-1$$, $$ d=- 4i+j-4k $$
Question 10
1 / -0

Four forces of magnitude P, 2P, 3P and 4P act along the four sides of a square ABCD in cyclic order. Use the vector method to find the resultant force.

A
$$\displaystyle 2\sqrt{2}P$$

B
$$\displaystyle 3\sqrt{2}P$$

C
$$\displaystyle \sqrt{2}P$$

D
$$\displaystyle -2\sqrt{2}P$$

Solution

The resultant of 4 vectors in X, Y direction are:
resultant in X direction is
$$3\vec P - \vec P = 2\vec P $$ (in -ve x direction)
In y direction is
$$4\vec P - 2\vec P = 2\vec P$$ (in +ve y direction)

So, the net resultant force vector is
$$\overset { \rightarrow }{ R } =2\overset { \rightarrow }{ P } +2\overset { \rightarrow }{ P } $$
angle between both vectors is 90
So magnitude of resultant is $$2\sqrt { 2 } P$$ in direction of $${ 45 }^{ 0 }$$ from positive y axis in IV quadrant.

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Vector Algebra Test - 37

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Vector Algebra Test - 37

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SHARING IS CARING

Question 1

$$2OE$$

$$OE$$

$$4OE$$

none of these

Solution

Question 2

$$(3/7) (i + 2j + 3k)$$

$$(i + 2j + 3k)$$

$$(1/7) (i + 2j + 3k)$$

$$(4/7) (i + 2j + 3k)$$

Solution

Question 3

$$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left ( \bar b-\bar a \right )$$,

$$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left ( \bar a-\bar b \right )$$,

$$\displaystyle \overrightarrow{AC}=\frac{2}{3}\left (\bar b+\bar a \right )$$,

$$\displaystyle \overrightarrow{AC}=\frac{3}{2}\left ( \bar b-\bar a \right )$$,

Solution

Question 4

Parallelogram

Trapezium

Rectangle

Both A and C

Solution

Question 5

$$(a\cdot i)i+(a\cdot j)j+(a\cdot k)k$$

$$(a\cdot j)i+(a\cdot k)j+(a\cdot i)k$$

$$(a\cdot k)j+(a\cdot i)j+(a\cdot j)k$$

$$(a\cdot a)i+(i+j+k) k$$

Solution

Question 6

$$\displaystyle 8/\sqrt{(35)}$$

$$\displaystyle 8/\sqrt{(39)}$$

$$\displaystyle 8/\sqrt{(14)}$$

$$\displaystyle \sqrt{(14)}$$

Solution

Question 7

$$\left (-\dfrac {20}{7}, -\dfrac {30}{7}, \dfrac {10}{7}\right)$$

$$(-20, -30, 10)$$

$$(2, 3, -1)$$

none of these

Solution

Question 8

$$\dfrac{3}{7}$$

$$\dfrac{7}{3}$$

$$3$$

$$7$$

Solution

Question 9

$$-4i+j-4k$$

$$3i+j-3k$$

$$ i+j-2k$$

$$4i+j-4k$$

Solution

Question 10

$$\displaystyle 2\sqrt{2}P$$

$$\displaystyle 3\sqrt{2}P$$

$$\displaystyle \sqrt{2}P$$

$$\displaystyle -2\sqrt{2}P$$

Solution

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