Vector Algebra Test

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Vector Algebra Test - 48

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Question 1
1 / -0

If $$\overline{a},\overline{b},\overline{c},$$ are unit vectors such that $$\overline{a}+\overline{b}+\overline{c}+\overline{c.a}=$$

A
$$\dfrac{3}{2}$$

B
$$-\dfrac{3}{2}$$

C
$$\dfrac{1}{2}$$

D
$$-\dfrac{1}{2}$$

Solution

$$-\dfrac{3}{2}$$
Question 2
1 / -0

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

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

B
$$6$$

C
$$-6$$

D
$$3$$

Solution

Projection of $$xi-j+k$$ on $$2i-j+5k$$
$$=\cfrac { \left( xi-j+k \right) \left( 2i-j+5k \right) }{ \sqrt { 4+1+25 } } =\cfrac { 2x+1+5 }{ \sqrt { 30 } } $$
But $$\cfrac { 2x+6 }{ \sqrt { 30 } } =\cfrac { 1 }{ \sqrt { 30 } }$$
$$ \Rightarrow x=-\cfrac { 5 }{ 2 } $$
Question 3
1 / -0

$$a$$ and $$c$$ are unit vectors and $$|b| = 4$$. If angle between $$b$$ and $$c$$ is $$\cos^{-1}\left (\dfrac {1}{4}\right )$$ and $$a\times b = 2a\times c$$, then $$b= \lambda a + 2c$$, where $$\lambda$$ is equal to

A
$$\pm \dfrac {1}{4}$$

B
$$\pm \dfrac {1}{2}$$

C
$$\pm 4$$

D
None of the above

Solution

Given, $$a\times b = 2a\times c$$
$$\Rightarrow a\times b - 2a\times c = 0$$
$$\Rightarrow a\times (b - 2c) = 0$$
$$\Rightarrow a$$ and $$(b - c)$$ are collinear.
$$\Rightarrow b - 2c = \lambda a$$
$$\Rightarrow |b - 2c|^{2} = \lambda^{2}|a|^{2}$$
$$\Rightarrow |b|^{2} + 4|c|^{2} - 4b\cdot c = \lambda^{2}|a|^{2}$$
$$\Rightarrow 16 + 4 - 4 (4) (1)\dfrac {1}{4} = \lambda^{2}(1)$$
Therefore, $$\lambda = \pm 4$$.
Question 4
1 / -0

Let $$\vec{OB} =\hat { i } +2\hat { j } +2\hat { k }$$ and $$\vec{OA} =4\hat { i } +2\hat { j } +2\hat { k } $$. The distance of the point $$B$$ from the straight line passing through $$A$$ and parallel to the vector $$2\hat { i } +3\hat { j } +6\hat { k } $$ is

A
$$\dfrac { 7\sqrt { 5 } }{ 9 } $$

B
$$\dfrac { 5\sqrt { 7 } }{ 9 } $$

C
$$\dfrac { 3\sqrt { 5 } }{ 7 } $$

D
$$\dfrac { 9\sqrt { 5 } }{ 7 } $$

E
$$\dfrac { 9\sqrt { 7 } }{ 5 } $$

Solution

Let the equation of line be $$r=a+\lambda b$$.
Since, line passes through $$A$$, so $$a=4\hat { i } +2\hat { j } +2\hat { k } $$ and line is parallel to the vector $$2\hat { i } +3\hat { j } +6\hat { k } $$, so $$b=2\hat { i } +3\hat { j } +6\hat { k } $$
Hence, equation of the line is
$$r=4\hat { i } +2\hat { j } +2\hat { k } +\lambda \left( 2\hat { i } +3\hat { j } +6\hat { k } \right) $$
Distance of point $$B$$ from the line $$=\left| \dfrac { \left| \left( { a }_{ 2 }-{ a }_{ 1 } \right) \right| \times b }{ \left| b \right| } \right| $$
$${ a }_{ 2 }-{ a }_{ 1 }=3\hat { i } +0\hat { j } +0\hat { k } $$
$$\left( { a }_{ 2 }-{ a }_{ 1 } \right) \times b=\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 0 & 0 \\ 2 & 3 & 6 \end{vmatrix}=-18\hat { j } +9\hat { k } $$
$$\left| \left( { a }_{ 2 }-{ a }_{ 1 } \right) \times b \right| =\sqrt { { \left( -18 \right) }^{ 2 }+{ \left( 9 \right) }^{ 2 } } $$
$$=\sqrt { 405 } =9\sqrt { 5 } $$
$$\left| b \right| =\sqrt { 4+9+36 } =7$$
Therefore, required distance is $$\dfrac { 9\sqrt { 5 } }{ 7 } $$.
Question 5
1 / -0

Given that A$$+$$B$$+$$C$$=0$$. Out of three vectors, two are equal in magnitude and the magnitude of third vector is $$\sqrt{2}$$ times that of either of the two having equal magnitude. Then, the angles between the vectors are given by.

A
$$30^o, 60^o, 90^o$$

B
$$45^o, 45^o, 90^o$$

C
$$45^o, 60^o, 90^o$$

D
$$90^o, 135^o, 135^o$$

Solution

From polygon law, three vectors having summation zero should form, a closed polygon(triangle). Since, the two vectors are having same magnitude and the third vector is $$\sqrt{2}$$ times that of either of two having equal magnitude, i.e., the triangle should be right angled triangle.
Angle between A and B, $$\alpha =90^o$$
Angle between B and C, $$\beta =135^o$$
Angle between A and C, $$\gamma =135^o$$.
Question 6
1 / -0

Let $$a,b,c$$ be three non-zero vectors such that no two of these are collinear. If the vectors $$a+2b$$ is collinear with $$c$$ and $$b+3c$$ is collinear with $$a$$ ($$\lambda$$ being some non-zero scalar), then $$a+2b+6c$$ equals to

A
$$\lambda a$$

B
$$\lambda b$$

C
$$\lambda c$$

D
$$0$$

Solution

Since, $$a+2b$$ is collinear with $$c$$
$$a+2b=mc...(i)$$
Since $$b+3c$$ is collinear with $$a$$
$$b+3c=na.....(ii)\quad $$
Myltiplying Eq(ii) by $$2$$ and subtracting Eq.(ii) from Eq.(i), we get
$$a-bc=mc-2na$$
On comparing, we get
$$m=-6;n=-\cfrac { 1 }{ 2 } $$
From Eq.(i), we have
$$a+2b=-6c\quad $$
$$\Rightarrow a+b+6c=0\quad $$
Question 7
1 / -0

Let $$\vec { a } =\hat { i } +\hat { j } -\hat { k } ,\vec { b } =\hat { i } -\hat { j } +\hat { k } $$ and $$\vec { c } $$ be a unit vector perpendicular to $$\vec { a } $$ and coplanar with $$\vec { a } $$ and $$\vec { b } $$, then $$\vec { c } $$ is

A
$$\cfrac { 1 }{ \sqrt { 2 } } \left( \hat { j } +\hat { k } \right) $$

B
$$\cfrac { 1 }{ \sqrt { 2 } } \left( \hat { j } -\hat { k } \right) $$

C
$$\cfrac { 1 }{ \sqrt { 6 } } \left( \hat { i } -2\hat { j } +\hat { k } \right) $$

D
$$\cfrac { 1 }{ \sqrt { 6 } } \left( 2\hat { i } -\hat { j } +\hat { k } \right) $$

Solution

Required unit vector $$=\pm \cfrac { \vec { a } \times \left( \vec { a } \times \vec { b } \right) }{ \left| \vec { a } \times \left( \vec { a } \times \vec { b } \right) \right| } $$
Now $$\vec { a } \times \left( \vec { a } \times \vec { b } \right) =\left( \vec { a } .\vec { b } \right)\vec{a} -\left( \vec { a } .\vec { a } \right) \vec { b } $$
$$\vec { a } \times \vec { b } =\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix}=-2\hat { j } -2\hat { k } $$
$$\therefore \vec { a } \times \left( \vec { a } \times \vec { b } \right) =\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & -1 \\ 0 & -2 & -2 \end{vmatrix}=-4\hat { i } +2\hat { j } -2\hat { k } $$
$$\left| \vec { a } \times \left( \vec { a } \times \vec { b } \right) \right| =\sqrt { 24 } =2\sqrt { 6 } $$
$$\therefore$$ Unit vector $$=\pm \cfrac { \vec { a } \times \left( \vec { a } \times \vec { b } \right) }{ \left| \vec { a } \times \left( \vec { a } \times \vec { b } \right) \right| } =\pm \cfrac { \left( 2\hat { i } -\hat { j } +\hat { k } \right) }{ \sqrt { 6 } } $$
(taking positive for option)
Question 8
1 / -0

Let $$a, b, c$$ be three non-coplanar vectors and $$r$$ be any arbitrary vector, then the expression $$(\vec {a}\times \vec {b})\times (\vec {r}\times \vec {c}) + (\vec {b} \times \vec {c})\times (\vec {r}\times \vec {a}) + (\vec {c}\times \vec {a}) \times (\vec {r}\times \vec {b})$$ is always equals to

A
$$[\vec {a}\vec {b}\vec {c}]\vec {r}$$

B
$$2[\vec {a}\vec {b}\vec {c}]\vec {r}$$

C
$$4[\vec {a}\vec {b}\vec {c}]\vec {r}$$

D
$$\vec {0}$$

Solution
Question 9
1 / -0

Find $$k$$ if magnitude of vectors joining $$(0,k,0)$$ and $$(1,1,1)$$ is $$\sqrt {11}$$

A
$$-3$$

B
$$2$$

C
$$-2$$

D
$$0$$

Solution

Using definition of magnitude, we have
$$11 = 1 + $$ $$(k-1)^{2}$$ $$+1$$
So $$(k-1)^{2}$$ $$= 9 $$
$$\Rightarrow k^2-2k+1-9=0$$
$$\Rightarrow k = +4$$ or $$-2 $$
So, the answer is $$ -2 $$.
Question 10
1 / -0

If three vectors $$ a, b, c $$ satisfy $$ a+b+c=0$$ and $$ |a| = 3, |b| = 5, |c| = 7 , $$ then the angle between $$a$$ and $$b$$ is :

A
$$30^o$$

B
$$45^o$$

C
$$60^o$$

D
$$90^o$$

Solution

Given, $$|a|=3,\quad |b|=5,\quad |c|=7$$ and
$$\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0$$
=> $$\overrightarrow { a } +\overrightarrow { b } =(-\overrightarrow { c } )$$
On squaring both side, we get
$${ (\overrightarrow { a } +\overrightarrow { b } ) }^{ 2 }={ (-\overrightarrow { c } ) }^{ 2 }$$
=> $${ |\overrightarrow { a } | }^{ 2 }+{ |\overrightarrow { b } | }^{ 2 }+2(\overrightarrow { a } .\overrightarrow { b } )={ |\overrightarrow { c } | }^{ 2 }$$
Let the angle between $$\overrightarrow { a } $$ and $$\overrightarrow { b } $$ is $$\theta $$, then
$${ |\overrightarrow { a } | }^{ 2 }+{ |\overrightarrow { b } | }^{ 2 }+2|\overrightarrow { a } ||\overrightarrow { b } |\cos { \theta } ={ |\overrightarrow { c } | }^{ 2 }$$
=> $${ 3 }^{ 2 }+{ 5 }^{ 2 }+2\times 3\times 5\times \cos { \theta } ={ 7 }^{ 2 }$$
=> $$\cos { \theta } =\frac { 1 }{ 2 } $$
So, the angle between $$\overrightarrow { a } $$ and $$\overrightarrow { b } $$ is $$\theta $$ $$= 60^o$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 48

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Vector Algebra Test - 48

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SHARING IS CARING

Question 1

$$\dfrac{3}{2}$$

$$-\dfrac{3}{2}$$

$$\dfrac{1}{2}$$

$$-\dfrac{1}{2}$$

Solution

Question 2

$$-\cfrac { 5 }{ 2 } $$

$$6$$

$$-6$$

$$3$$

Solution

Question 3

$$\pm \dfrac {1}{4}$$

$$\pm \dfrac {1}{2}$$

$$\pm 4$$

None of the above

Solution

Question 4

$$\dfrac { 7\sqrt { 5 } }{ 9 } $$

$$\dfrac { 5\sqrt { 7 } }{ 9 } $$

$$\dfrac { 3\sqrt { 5 } }{ 7 } $$

$$\dfrac { 9\sqrt { 5 } }{ 7 } $$

$$\dfrac { 9\sqrt { 7 } }{ 5 } $$

Solution

Question 5

$$30^o, 60^o, 90^o$$

$$45^o, 45^o, 90^o$$

$$45^o, 60^o, 90^o$$

$$90^o, 135^o, 135^o$$

Solution

Question 6

$$\lambda a$$

$$\lambda b$$

$$\lambda c$$

$$0$$

Solution

Question 7

$$\cfrac { 1 }{ \sqrt { 2 } } \left( \hat { j } +\hat { k } \right) $$

$$\cfrac { 1 }{ \sqrt { 2 } } \left( \hat { j } -\hat { k } \right) $$

$$\cfrac { 1 }{ \sqrt { 6 } } \left( \hat { i } -2\hat { j } +\hat { k } \right) $$

$$\cfrac { 1 }{ \sqrt { 6 } } \left( 2\hat { i } -\hat { j } +\hat { k } \right) $$

Solution

Question 8

$$[\vec {a}\vec {b}\vec {c}]\vec {r}$$

$$2[\vec {a}\vec {b}\vec {c}]\vec {r}$$

$$4[\vec {a}\vec {b}\vec {c}]\vec {r}$$

$$\vec {0}$$

Solution

Question 9

$$-3$$

$$2$$

$$-2$$

$$0$$

Solution

Question 10

$$30^o$$

$$45^o$$

$$60^o$$

$$90^o$$

Solution

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