Vector Algebra Test

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Vector Algebra Test - 46

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Question 1
1 / -0

If the scalar projection of the vector $$xi-j+k$$ on the vector $$2i-j+5k$$ is $$\dfrac { 1 }{ \sqrt { 30 } } $$ then value of $$x$$ is equal to

A
$$\dfrac { -5 }{ 2 } $$

B
$$6$$

C
$$-6$$

D
$$3$$

Solution

Scalar projection of $$\vec { a } $$ on $$\vec { b } $$ $$=\dfrac { \left| \vec { a } .\vec { b } \right| }{ \left| \vec { b } \right| } $$

let $$ \vec { a } \ =xi-j+k$$ and $$\vec { b } =2i-j+5k$$

Then projection of $$\vec { a } $$ on $$\vec { b } $$ $$=\dfrac { (xi-j+k).(2i-j+5k) }{ \left| 2i-j+5k \right| } $$

Given scalar projection $$=\dfrac { 1 }{ \sqrt { 30 } } $$

$$\Rightarrow \dfrac { 1 }{ \sqrt { 30 } } =\dfrac { 2x+1+5 }{ \sqrt { { 2 }^{ 2 }+{ (-1) }^{ 2 }+{ 5 }^{ 2 } } } \\ \Rightarrow \dfrac { 1 }{ \sqrt { 30 } } =\dfrac { 2x+1+5 }{ \sqrt { 30 } } \\ \Rightarrow 2x+1+5=1\\ \Rightarrow x=-\dfrac { 5 }{ 2 } $$
Question 2
1 / -0

Let $$\square PQRS$$ be a quadrilateral. If $$M$$ and $$N$$ are midpoints of the sides $$PQ$$ and $$RS$$ respectively then $$\overline {PS} + \overline {QR} =$$

A
$$3\overline {MN}$$

B
$$4\overline {MN}$$

C
$$2\overline {MN}$$

D
$$2\overline {NM}$$

Solution

Let $$\overline{m}, \overline{n}, \overline{p}, \overline{q}, \overline{r}$$ and $$\overline{s}$$ be position vectors of $$M,N,P,Q,R$$ and $$S$$ respectively.

$$M$$ and $$N$$ are midpoints of $$PQ$$ and $$RS$$
$$\therefore \overline {m} = \dfrac {\overline {p} + \overline {q}}{2}$$ and
$$\overline {n} = \dfrac {\overline {r} + \overline {s}}{2}$$

$$\overline {PS} + \overline {QR} = \overline {s} - \overline {p} + \overline {r} - \overline {q}$$
$$= (\overline {r} + \overline {s}) - (\overline {p} + \overline {q})$$
$$= 2\overline {n} - 2\overline {m}$$

$$= 2(\overline {n} - \overline {m})$$

$$= 2\overline {MN}$$
Question 3
1 / -0

If $$\vec{a}$$ and $$\vec{b}$$ are the vectors determined by two adjacent sides of regular hexagon, then vector $$EF$$ is

A
$$(\vec{a} + \vec{b})$$

B
$$(\vec{a} - \vec{b})$$

C
$$2 \vec{a}$$

D
$$2\vec{b}$$

Solution

FABCDE is a regular hexgon.

Let $$\overline{FA}=\vec{a}$$ and $$\overline{AB}=\vec{b}$$, join $$FB$$ and $$FC$$, we have
$$\overline{FB}=\overline{FA}+\overline{AB}=\vec{a}+\vec{b}$$

Since, $$\overline{FC}$$ is parallel to $$\overline{AB}$$ and double of

$$\overline{AB}$$
$$\therefore \overline{FC}=2\overline{AB}=2\vec{b}$$

Now, $$\overline{BC}=FC-FB$$$$=2\vec{b}-(\vec{a}+\vec{b})=\vec{b}-\vec{a}$$

$$\overline{CD}=-\overline{FA}=-\vec{a}$$ and $$\overline{DE}=-\overline{AB}=-\vec{b}$$

Also, $$\overline{EF}=-\overline{BC}=-(\vec{b}-\vec{a})=\vec{a}-\vec{b}$$
Question 4
1 / -0

If $$p=\hat {i} + \hat{j}, q=4\hat{k}-\hat{j}$$ and $$r=\hat{i}+\hat{k}$$, then the unit vector in the direction of $$3p+q-2r$$ is

A
$$\dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$$

B
$$\dfrac{1}{3}(\hat{i}-2\hat{j}-2\hat{k})$$

C
$$\dfrac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})$$

D
$$\hat{i}-2\hat{j}+2\hat{k}$$

Solution

$$3p+q-2r=3(\hat{i}+\hat{j})+(4\hat{k}-\hat{j})-2(\hat{i}+\hat{k})$$

$$=\hat{i}+2\hat{j}+2\hat{k}$$

$$\therefore$$ Unit vector in the direction of $$3p+q-2r = \dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$$
Question 5
1 / -0

If $$\overrightarrow a$$ and $$\overrightarrow b$$ are unit vectors, then angle between $$\overrightarrow a$$ and $$\overrightarrow b$$ for $$\sqrt 3 \overrightarrow a - \overrightarrow b$$ to be unit vector is

A
$$60^o$$

B
$$45^o$$

C
$$30^o$$

D
$$90^o$$

Solution

$${\textbf{Step-1: Apply unit vector property.}}$$
$${\text{Given,}}$$
$$ \bar a$$ $${\text{and}}$$ $$\bar b$$ $${\text{are unit vectors and}}$$ $$( \sqrt 3 \bar a - \bar b)$$ $${\text{is also unit vector.}}$$
$${\text{Let angle be}}$$ $$\theta$$
$$\Rightarrow \bar a. \bar b =| \bar a| | \bar b| cos \theta$$ $${\text{.....Dot product of two vectors}}$$
$${\text{But,}}$$ $$\bar a$$ $${\text{and}}$$ $$ \bar b $$ $${\text{are unit vectors.}}$$
  $$|\bar a |=$$  $$| \bar b |= 1$$
  $$\Rightarrow \bar a. \bar b =cos \theta$$ $$....(1)$$
  $$( \sqrt 3 \bar a - \bar b) =1$$  $${\text{also unit vector.}}$$
$${\textbf{Step-2: Squaring on both side and find angle.}}$$
  $$( \sqrt 3 \bar a - \bar b)^2 =1$$
  $$( \sqrt 3 )^2 | \bar a|^2 - 2 \sqrt3 |\bar a. \bar b| + | \bar b|^2 =1$$ $$.......\because (a-b)^2 = a^2-2ab +b^2$$
$$\Rightarrow 3(1) + (1) - 2 \sqrt3 cos \theta = 1$$ $$....(1)$$
$$\Rightarrow 4 - 2 \sqrt3 cos \theta = 1$$
$$\Rightarrow 2 \sqrt3 cos \theta = 3$$
$$\Rightarrow cos \theta = \dfrac{\sqrt3}{2} = cos 30 ^\circ$$
$$\Rightarrow \theta = 30 ^\circ$$
$${\textbf{Hence, the option (C) is correct.}}$$
Question 6
1 / -0

If $$a.b=0$$ and $$a+b$$ makes an angle of $${ 60 }^{ o }$$ with $$b$$, then $$\left| a \right| $$ is equal to

A
$$0$$

B
$$\cfrac { 1 }{ \sqrt { 3 } } \left| b \right| $$

C
$$\cfrac { 1 }{ \left| b \right| } $$

D
$$\left| b \right| $$

E
$$\sqrt { 3 } \left| b \right| $$

Solution

We have $$a.b=0\Rightarrow a\bot b$$
so, vectors $$a,b$$ and $$a+b$$ form a right angled triangle
In $$\triangle PQR\quad $$
$$\tan { { 60 }^{ o } } =\cfrac { \left| a \right| }{ \left| b \right| } $$
$$\Rightarrow \sqrt { 3 } =\cfrac { \left| a \right| }{ \left| b \right| } $$
$$\Rightarrow \left| a \right| =\sqrt { 3 } \left| b \right| $$
Question 7
1 / -0

If $$a\cdot b =0$$ and $$a + b$$ makes an angle $$60^o$$ with $$a$$, then

A
$$|a| = 2 |b|$$

B
$$2 |a| = |b|$$

C
$$|a| = \sqrt 3 |b|$$

D
$$|a| = |b|$$

E
$$\sqrt 3 |a | = |b|$$

Solution

Given, $$a\cdot b = 0$$
Now, $$(a + b)\cdot a = |a + b||a| \cos 60^o$$
$$\Rightarrow a\cdot a + b\cdot a = |a + b| |a| \dfrac{1}{2}$$
$$\Rightarrow |a|^2 + 0 = \dfrac{|a + b| |a|}{2}$$
$$\Rightarrow 2 |a| = |a + b|$$
On squaring both sides, we get
$$4 |a|^2 = |a|^2 + |b|^2 + 2 a\cdot b \cos \theta$$
$$4 |a|^2 = |a|^2 + |b|^2 + 2 |a| |b| \cos \theta$$
$$\Rightarrow 3 |a|^2 = |b|^2 + 0 (\because |a| |b| \cos \theta = a \cdot b = 0 )$$
$$\Rightarrow \sqrt 3 |a| = |b|$$
Question 8
1 / -0

If $$a\times b=c$$, $$b\times c=a$$ and $$a,b,c$$ be the mod of the vectors $$a,b,c$$ respectively, then

A
$$a=1, b=1$$

B
$$c=1, a=1$$

C
$$a\cdot \left( b\times c \right) =1$$

D
$$b=1,c=a$$

Solution

Since, $$a\times b=c$$
Thus $$ \left( b\times c \right) \times b=c$$
$$\Rightarrow \left( b\cdot b \right) c-\left( b\cdot c \right) b=c$$
$$\Rightarrow \left( b\cdot b \right) =1$$, $$c\cdot b=0$$
$$\Rightarrow { b }^{ 2 }=1$$ i.e., $$b=1$$
$$ b$$ is a unit vector
Therefore, $$ \left| c \right| =\left| a \right| \Rightarrow c=a$$
Question 9
1 / -0

If $$\widehat i + \widehat j, \widehat j + \widehat k, \widehat i + \widehat k$$ are the position vectors of the vertices of a $$\Delta ABC$$ taken in order, then $$\angle A$$ is equal to

A
$$\dfrac {\pi}{2}$$

B
$$\dfrac {\pi}{5}$$

C
$$\dfrac {\pi}{6}$$

D
$$\dfrac {\pi}{4}$$

E
$$\dfrac {\pi}{3}$$

Solution

Let position vector of the vertices are
$$OA = \widehat i + \widehat j, OB = \widehat j + \widehat k$$
and $$OC = \widehat i + \widehat k$$
Now, $$AB = - \widehat i + \widehat k$$
and $$AC = \widehat k - \widehat j$$
$$\therefore \cos \theta = \dfrac{(AB).(AC)}{|AC||AC|}$$
$$= \dfrac{(-\widehat i + \widehat k) . (\widehat k - \widehat j)}{\sqrt{1^2 + 1^2} \sqrt{1^2 + 1^2}}$$
$$= \dfrac{(\widehat k)^2}{\sqrt 2 \sqrt 2} = \dfrac{1}{2}$$
$$\Rightarrow \theta = \dfrac{\pi}{3}$$
Question 10
1 / -0

In the given diagram, if PQ$$=$$A, QR$$=$$B and RS$$=$$C, then PS will be equal to :

A
$$A-B+C$$

B
$$A+B-C$$

C
$$A+B+C$$

D
$$A-B-C$$

E
$$-A-B-C$$

Solution

According to polygon law of vector addition
$$A+B=PR$$
and $$PR+RS=PS$$
$$\Rightarrow A+B+C=PS$$.

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Vector Algebra Test - 46

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Vector Algebra Test - 46

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SHARING IS CARING

Question 1

$$\dfrac { -5 }{ 2 } $$

$$6$$

$$-6$$

$$3$$

Solution

Question 2

$$3\overline {MN}$$

$$4\overline {MN}$$

$$2\overline {MN}$$

$$2\overline {NM}$$

Solution

Question 3

$$(\vec{a} + \vec{b})$$

$$(\vec{a} - \vec{b})$$

$$2 \vec{a}$$

$$2\vec{b}$$

Solution

Question 4

$$\dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$$

$$\dfrac{1}{3}(\hat{i}-2\hat{j}-2\hat{k})$$

$$\dfrac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})$$

$$\hat{i}-2\hat{j}+2\hat{k}$$

Solution

Question 5

$$60^o$$

$$45^o$$

$$30^o$$

$$90^o$$

Solution

Question 6

$$0$$

$$\cfrac { 1 }{ \sqrt { 3 } } \left| b \right| $$

$$\cfrac { 1 }{ \left| b \right| } $$

$$\left| b \right| $$

$$\sqrt { 3 } \left| b \right| $$

Solution

Question 7

$$|a| = 2 |b|$$

$$2 |a| = |b|$$

$$|a| = \sqrt 3 |b|$$

$$|a| = |b|$$

$$\sqrt 3 |a | = |b|$$

Solution

Question 8

$$a=1, b=1$$

$$c=1, a=1$$

$$a\cdot \left( b\times c \right) =1$$

$$b=1,c=a$$

Solution

Question 9

$$\dfrac {\pi}{2}$$

$$\dfrac {\pi}{5}$$

$$\dfrac {\pi}{6}$$

$$\dfrac {\pi}{4}$$

$$\dfrac {\pi}{3}$$

Solution

Question 10

$$A-B+C$$

$$A+B-C$$

$$A+B+C$$

$$A-B-C$$

$$-A-B-C$$

Solution

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