Vector Algebra Test

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Vector Algebra Test - 69

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Question 1
1 / -0

$$EFGH$$ is a rhombus such the angle $$EFG$$ is $${60}^{o}$$. The magnitude of vectors $$\overrightarrow { FH } $$ and $$\left\{ m\overrightarrow { EG } \right\} $$ are equal where $$m$$ is a scalar. What is the value of $$m$$?

A
$$3$$

B
$$1.5$$

C
$$\sqrt { 2 } $$

D
$$\sqrt { 3 } $$

Solution

Let the side of rhombus
$$\vec{EF}=\hat{i}+\hat{j}$$
$$\vec{FG}=\hat{j}+\hat{k}$$
Here the angle between $$\vec{EF},\vec{FG}$$ is 60
$$\vec{EF}\cdot\vec{FG}=\left | \vec{EF} \right |\left | \vec{FG} \right |\cos\theta$$
$$(\hat{i}+\hat{j})\cdot(\hat{j}+\hat{k})=\sqrt{1^2+1^2}\sqrt{1^2+1^2}\cos\theta$$
$$1=\sqrt{2}\sqrt{2}\cos\theta$$
$$\dfrac{1}{2}=\cos\theta$$
$$\theta=60$$
Hence our vector satisfying the given condition

EG is the diagonal so by triangle law in triangle EFG
$$\vec{EG}=\vec{EF}+\vec{FG}$$
$$\vec{EG}=\hat{i}+\hat{j}+\hat{j}+\hat{k}$$
$$\vec{EG}=\hat{i}+2\hat{j}+\hat{k}$$
FH is also diagonal so in triangle FGH
$$\vec{FH}=\vec{FG}+\vec{GH}$$
but $$\vec{GH}=-\vec{EF}$$
$$\vec{FH}=\vec{FG}-\vec{EF}$$
$$\vec{FH}=\hat{j}+\hat{k}-\hat{i}-\hat{j}$$
$$\vec{FH}=-\hat{i}+\hat{k}$$
$$m\left | \vec{FH} \right |=\left | \vec{EG} \right |$$
$$m\sqrt{(-1)^2+1^2}=\sqrt{1^2+2^2+1^2}$$
$$m\sqrt{1+1}=\sqrt{1+4+1}$$
$$m\sqrt{2}=\sqrt{6}$$
$$m=\dfrac{\sqrt{6}}{\sqrt{2}}$$
$$m=\sqrt{3}$$
Question 2
1 / -0

The direction cosines l , m and n of two lines are connected by the relations l+m+n=0, lm=0, then the angles between them is

A
$$\displaystyle \pi \over 3$$

B
$$\displaystyle \pi \over 4$$

C
$$\displaystyle \pi \over 2$$

D
$$0$$

Solution
Question 3
1 / -0

If the sum of two unit vectors is also a unit vector, then the angle between the two vectors is

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

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

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

D
None of these

Solution

Given condition
$$\vec { a } +\vec { b } =\vec { c } $$
where
$$\left| \vec { a } \right| =\left| \vec { b } \right| =\left| \vec { c } \right| =1$$
Squaring both sides
$${ \left| \vec { a } \right| }^{ 2 }+{ \left| \vec { b } \right| }^{ 2 }+2.\left| \vec { a } \right| .\left| \vec { b } \right| .\cos { \theta } ={ \left| \vec { c } \right| }^{ 2 }$$
$$1+1+2\cos { \theta } ={ \left| \vec { c } \right| }^{ 2 }=1$$
$$2\cos { \theta } =-1$$
$$\cos { \theta } =\cfrac { -1 }{ 2 } $$
$$\cos { \theta } =\cos { \left( \pi -\cfrac { \pi }{ 3 } \right) } $$
$$\theta =\cfrac { 2\pi }{ 3 } $$
Question 4
1 / -0

If P(x, y, z) is a point on the line segment joining Q(2, 2, 4) and R(3, 5, 6) such that the projections of $$\overrightarrow{OP}$$ on the axes are $$\dfrac{13}{5}, \dfrac{19}{5}, \dfrac{26}{5}$$ respectively, then P divides QR in the ratio

A
1 : 2

B
3 : 2

C
2 : 3

D
1 : 3

Solution

Let $$ P \left ( x,y,z \right ) $$
$$ Q \left ( 2,2,4 \right ) $$
$$ R \left ( 3,5,6 \right ) $$

Let the ratio in which $$ P \left ( x,y,z \right ) $$ divides the line joining $$ Q \left ( 2,2,4 \right ), R \left ( 3,5,6 \right ) $$ be $$\lambda :1 $$

According to the section formula ,
$$P\left ( x,y,z \right ) =\left ( \dfrac{3\lambda +2}{\lambda +1} ,\dfrac{5\lambda +2}{\lambda +1},\dfrac{6\lambda +4}{\lambda +1}\right )$$

Projection of $$ \vec{a} $$ on $$\vec{b}$$ is $$ \dfrac{\vec{a}\cdot \vec{b}}{ \left | \vec{b} \right |} $$
It is given that projection of $$ \vec{OP} $$ on coordinate axes are $$\dfrac {13}{5}, \dfrac {19}{5}, \dfrac{26}{5} $$ is
$$ \dfrac{\left ( x,y,z \right )\cdot \left ( 1,0,0 \right )}{ \left | 1,0,0 \right |} = \dfrac{13}{5} $$

$$ x= \dfrac{13}{5} = \dfrac{3\lambda +2}{\lambda +1} $$

$$\lambda =\dfrac{3}{2} $$
Question 5
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What are coinitial vectors.?

A
Two or more vectors having the same magnitude are called coinitial vectors.

B
Two or more vectors are said to be coinitial if they are parallel to the same line, irrespective of their magnitudes and directions.

C
Two or more vectors having the same initial point are called coinitial vectors.

D
None of the above

Solution

two vectors having same magnitude called equal vector
Hence option $$A$$ is not correct
option $$B$4 also is the property of equal vector
coinitail means the initail point is same
So option $$C$$ is correct
Question 6
1 / -0

If three non-zero vectors are $$a=a_1i+a_2j+a_3k , \, b=b_1i+b_2j+b_3k$$ and $$ c=c_1i+c_2j+c_3k$$ . If c is the unit vector perpendicular to the vectors a and b the angle between a and b is $$\dfrac{\pi}{6}$$ , then $$\begin{vmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}$$ is equal to

A
0

B
$$ \dfrac{3(\sum a^2_1)(\sum b^2_1) \sum c^2_1}{4}$$

C
1

D
$$\dfrac{(\sum a^2_1)(\sum b^2_1)}{4}$$
Question 7
1 / -0

If the vectors $$\overline { c } ,\overline { a } =x\hat { i } +y\hat { j } +z\hat { k } $$, and $$\overline { b } =\hat { j }$$ are such that $$\overline { a } ,\overline { c } \ and \ \overline { b }$$ from a right handed system, then $$\overline { c }$$ is

A
$$z\hat { i-x } \hat { k }$$

B
$$\overline { 0 }$$

C
$$y\hat { j }$$

D
$$-z\hat { i+x } \hat { k }$$

Solution
Question 8
1 / -0

Let $$O$$ be an interior point of $$\triangle ABC$$ such that $$\overline {OA} + 2\overline {AB} + 3\overline {OC} = \overline {O}$$, then the ratio of $$\triangle ABC$$ to area of $$\triangle AOC$$ is

A
$$2$$

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

C
$$3$$

D
$$\dfrac {5}{2}$$

Solution

Refer to the figure. Let O be the origin.
Let $$\bar { OA } =\bar { a } $$
$$\bar { OB } =\bar { b } $$
$$\bar { OC } =\bar { c } $$

Given, $$\bar { OA } +2\bar { OB } +3\bar { OC } =0$$

$$\therefore \bar { a } +2\bar { b } +3\bar { c } =0$$

$$\therefore \bar { a } =-2\bar { b } -3\bar { c } $$ (1)

1) Area of triangle ABC is given by,
$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \bar { AB } \times \bar { BC } \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } -\bar { a } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

From equation (1),

$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } -\left( -2\bar { b } -3\bar { c } \right) \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } +2\bar { b } +3\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( 3\bar { b } +3\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } +\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) -\left( \bar { b } \times \bar { b } \right) +\left( \bar { c } \times \bar { c } \right) -\left( \bar { c } \times \bar { b } \right) \right| $$

But, $$\bar { b } \times \bar { b } =0$$ and $$\bar { c } \times \bar { c } =0$$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) -\left( \bar { c } \times \bar { b } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) +\left( \bar { b } \times \bar { c } \right) \right| $$
$$\left( \because \left( \bar { a } \times \bar { b } \right) =-\left( \bar { b } \times \bar { a } \right) \right) $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| 2\left( \bar { b } \times \bar { c } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =3\left| \left( \bar { b } \times \bar { c } \right) \right| $$ (2)

2) Area of triangle AOC is given by,
$$\therefore \left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \bar { OA } \times \bar { OC } \right| $$

$$\therefore \left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \bar { a } \times \bar { c } \right| $$

From equation (1),
$$A\left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \left( -2\bar { b } -3\bar { c } \right) \times \bar { c } \right| $$

$$\therefore A\left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| -2\left( \bar { b } \times \bar { c } \right) -3\left( \bar { c } \times \bar { c } \right) \right| $$

$$\therefore A\left( \triangle AOC \right) =\frac { 2 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) \right| $$

$$\therefore A\left( \triangle AOC \right) =\left| \left( \bar { b } \times \bar { c } \right) \right| $$ (3)

Now, $$\frac { A\left( \triangle ABC \right) }{ A\left( \triangle AOC \right) } =\frac { 3\left| \left( \bar { b } \times \bar { c } \right) \right| }{ \left| \left( \bar { b } \times \bar { c } \right) \right| } $$

$$\frac { A\left( \triangle ABC \right) }{ A\left( \triangle AOC \right) } =3$$
Question 9
1 / -0

If vectors $$i+2j+2k$$ is rotated through an angle of $${90}^{o}$$ so as to cross positive direction of y-axis, then the vector in the new positive is?

A
$$-\cfrac { 2 }{ \sqrt { 5 } } i+\sqrt { 5 } j-\cfrac { 4 }{ \sqrt { 5 } } k$$

B
$$\cfrac { 2 }{ \sqrt { 5 } } i-\sqrt { 5 } j+\cfrac { 4 }{ \sqrt { 5 } } k$$

C
$$4i-j-k$$

D
None of these

Solution

Let the new position of the vector be $$OP′=xi+yj+zk.$$ As $$OP$$ is making $$90^{\circ}$$ with $$OP′.$$
So $$(xi+yj+zk).(i+2j+2k)=0⇒x+2y+2z=0 -----(1)$$
Also $$|\vec{OP}|=|\vec{OP′}|$$
$$x^2+y^2+z^2=9 ------(2)$$
Now if we say that the vector moves in such a way that it touches the y axis then we (perhaps) may assume that it moves in the plane determined by $$i$$ and $$i+2j+2k$$, the vector itself. A normal vector to this plane is $$j ×(i +2j +2k )=2i-k $$. The equation of such a plane is $$2x=z.$$
So we have the following equation from (1) and (2)
$$5x+2y=0$$
and $$5x^2+y^2=9$$
From here,
$$x=\pm\dfrac{2}{\sqrt{5}}$$
$$\rightarrow y=\mp\sqrt{5}$$
$$z=\pm\dfrac{4}{\sqrt{5}}$$
So the two vectors are $$OP'=\dfrac{2}{\sqrt{5}}i-\sqrt{5}j+\dfrac{4}{\sqrt{5}}k$$
and $$OP'=-\dfrac{2}{\sqrt{5}}i+\sqrt{5}j-\dfrac{4}{\sqrt{5}}k$$
Note that there are two solutions in general. But one of them does not touch the y axis while moving along the simplest path. Our solution is
$$OP'=-\dfrac{2}{\sqrt{5}}i+\sqrt{5}j-\dfrac{4}{\sqrt{5}}k$$
Hence, $$(A)$$
Question 10
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A point $$C = \dfrac {5\overline {a} + 4\overline {b} - 5\overline {c}}{3}$$ divides the line joining the points $$A$$ and $$B = 2\overline {a} + 3\overline {b} - 4\overline {c}$$ in the ratio $$2 : 1$$, then the position vector of $$A$$ is

A
$$\overline {a} + 3\overline {b} - 4\overline {c}$$

B
$$2\overline {a} - 3\overline {b} + 4\overline {c}$$

C
$$2\overline {a} + 3\overline {b} + 4\overline {c}$$

D
$$\overline {a} - 2\overline {b} + 3\overline {c}$$

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Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 69

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Vector Algebra Test - 69

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SHARING IS CARING

Question 1

$$3$$

$$1.5$$

$$\sqrt { 2 } $$

$$\sqrt { 3 } $$

Solution

Question 2

$$\displaystyle \pi \over 3$$

$$\displaystyle \pi \over 4$$

$$\displaystyle \pi \over 2$$

$$0$$

Solution

Question 3

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

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

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

None of these

Solution

Question 4

1 : 2

3 : 2

2 : 3

1 : 3

Solution

Question 5

Two or more vectors having the same magnitude are called coinitial vectors.

Two or more vectors are said to be coinitial if they are parallel to the same line, irrespective of their magnitudes and directions.

Two or more vectors having the same initial point are called coinitial vectors.

None of the above

Solution

Question 6

0

$$ \dfrac{3(\sum a^2_1)(\sum b^2_1) \sum c^2_1}{4}$$

1

$$\dfrac{(\sum a^2_1)(\sum b^2_1)}{4}$$

Question 7

$$z\hat { i-x } \hat { k }$$

$$\overline { 0 }$$

$$y\hat { j }$$

$$-z\hat { i+x } \hat { k }$$

Solution

Question 8

$$2$$

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

$$3$$

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

Solution

Question 9

$$-\cfrac { 2 }{ \sqrt { 5 } } i+\sqrt { 5 } j-\cfrac { 4 }{ \sqrt { 5 } } k$$

$$\cfrac { 2 }{ \sqrt { 5 } } i-\sqrt { 5 } j+\cfrac { 4 }{ \sqrt { 5 } } k$$

$$4i-j-k$$

None of these

Solution

Question 10

$$\overline {a} + 3\overline {b} - 4\overline {c}$$

$$2\overline {a} - 3\overline {b} + 4\overline {c}$$

$$2\overline {a} + 3\overline {b} + 4\overline {c}$$

$$\overline {a} - 2\overline {b} + 3\overline {c}$$

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