Vector Algebra Test

Result

Vector Algebra Test - 71

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The vector $$(\hat {i}\times \vec {a}.\vec {b})\hat {i} + (\hat {j} \times \vec {a}.\vec {b})\hat {j} + (\hat {k} \times \vec {a} . \vec {b})\hat {k}$$ is equal to

A
$$\vec {b}\times \vec {a}$$

B
$$\vec {a}$$

C
$$\vec {a}\times \vec {b}$$

D
$$\vec {b}$$
Question 2
1 / -0

The x-y plane divides the line joining the points $$(-1, 3, 4)$$ and $$(2, -5, 6)$$:

A
internally in the ratio $$2:3$$

B
externally in the ratio $$2:3$$

C
internally in the ratio $$3:2$$

D
externally in the ratio $$3:2$$

Solution

Let $$xy$$ plane divides the line joining the points $$A\left( { - 1,3,4} \right)$$ and $$B\left( {2, - 5,6} \right)$$ in the ratio $$k:1$$.
$$x = \cfrac{{2k - 1}}{{k + 1}}$$
$$y = \cfrac{{ - 5k + 3}}{{k + 1}}$$
$$z = \cfrac{{6k - 4}}{{k + 1}}$$
Since the plane is $$xy$$ plane, so $$z = 0$$,
$$\cfrac{{6k - 4}}{{k + 1}} = 0$$
$$6k - 4 = 0$$
$$6k = 4$$
$$k = \cfrac{4}{6}$$
$$k = \cfrac{2}{3}$$
Therefore, the ratio is $$2:3$$ internally.
Question 3
1 / -0

If $$\vec {a},\vec {b}$$ and $$\vec {c}$$ are those mutually perpendicular vectors, then the projection of the vector $$\left( l \dfrac{\bar {a}}{|\bar {a}|}+m\dfrac{\bar {b}}{|\bar {b}|}+n\dfrac{(\bar {a}\times \bar {b})}{|\bar {a} \times \bar {b}|}\right)$$ along bisector of vectors $$\vec {a}$$ and $$\vec {a}$$ may be given as ?

A
$$\dfrac{l^{2}+m^{2}}{\sqrt{l^{2}+m^{2}+n^{2}}}$$

B
$$\sqrt{l^{2}+m^{2}+n^{2}}$$

C
$$\dfrac{\sqrt{l^{2}+m^{2}}}{\sqrt{l^{2}+m^{2}+n^{2}}}$$

D
$$\dfrac{l+m}{\sqrt{2}}$$

Solution

Question :- If $$\vec{a},\vec{b}$$ and $$\vec{c}$$ are three mutually perpendicular
vectors, then projection of vector $$\left ( l\dfrac{\vec{a}}{\left | \vec{a} \right |}+m\dfrac{\vec{b}}{\left | \vec{b} \right |}+n\dfrac{(\vec{a}\times \vec{b})}{\left | \vec{a}\times \vec{b} \right |} \right )$$
along bisector of sectors $$\vec{a}$$ and $$\vec{b}$$ may be given as?
Solution:-
We know for mutually perp. vectors.
$$\vec{a}\times \vec{b}=\vec{c}$$ $$\vec{b}\times \vec{c}=\vec{a}$$ $$\vec{c}\times \vec{a}=\vec{b}$$
$$\vec{a}.\vec{b}=\vec{b}.\vec{c}=\vec{c}.\vec{a}=0$$ and $$\left | \hat{a} \right |^{2}=1$$
Now, A vector parallel to bisector of angle between
vectors $$\vec{a}$$ and $$\vec{b}$$ is given as $$\vec{x}$$
$$\vec{x}=\dfrac{\vec{a}}{\left | \vec{a} \right |}+\dfrac{\vec{b}}{\left | \vec{b} \right |}=\hat{a}+\hat{b}$$ $$(\hat{a}=\hat{b}$$ = unit vectors)
say $$\vec{y}=\left ( l\dfrac{\vec{a}}{\left | \vec{a} \right |} +m\dfrac{\vec{b}}{\left | \vec{b} \right |}+n\dfrac{(\vec{a}\times \vec{b})}{\left | \vec{a}\times \vec{b} \right |}\right )$$
$$\vec{y}=\left ( l\hat{a}+m\hat{b}+n\dfrac{\vec{c}}{\left | \vec{c} \right |} \right )$$
$$\vec{y}=(l\hat{a}+m\hat{b}+n\hat{c})$$
Now projection of y along x is given as
$$=\vec{y}.\dfrac{\vec{x}}{\left | \vec{x} \right |}$$
$$=\vec{y}.\hat{x}$$
$$=(l\hat{a}+m\hat{b}+n\hat{c}).\left ( \dfrac{\hat{a}}{\sqrt{2}}+\dfrac{\hat{b}}{\sqrt{2}} \right )$$
$$=\dfrac{l}{\sqrt{2}}+\dfrac{m}{\sqrt{2}}$$
$$=\dfrac{l+m}{\sqrt{2}}$$
Ans (D) $$\dfrac{l+m}{\sqrt{2}}$$

Using $$\vec{x}=\hat{a}+\hat{b}$$
$$\hat{x}=\dfrac{\hat{a+\hat{b}}}{\left | \hat{a}+\hat{b} \right |}=\dfrac{1}{\sqrt{2}}(\hat{a}+\hat{b})$$
$$\left | \hat{a}+\hat{b} \right |^{2}=\left | \hat{a} \right |^{2}+\left | \hat{b} \right |^{2}+2\hat{a}.\hat{b}$$
$$= 1+1+0=2$$
Question 4
1 / -0

$$\bar{a}, \bar{b}, \bar{c}$$ are mutually perpendicular unit vectors and $$\bar{d}$$ is a unit vector equally inclined to each other of $$\bar{a}, \bar{b}$$ and $$\bar{c}$$ at an angle of $$60^o$$. Then $$|\bar{a}+\bar{b}+\bar{c}+\bar{d}|^2=?$$

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

$${\textbf{Step -1: Solve the given equations}}{\textbf{.}}$$
$$\left| {\overline a } \right| = \left| {\overline b } \right| = \left| {\overline c } \right| = \left| {\overline d } \right| = 1$$
$${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + {\left| {\overrightarrow d } \right|^2} + 2\left( {\overrightarrow {a.} \overrightarrow b + \overrightarrow {a.} \overrightarrow {c.} + \overrightarrow {a.} \overrightarrow {d.} + \overrightarrow {b.} \overrightarrow {c.} + \overrightarrow {b.} \overrightarrow {a.} + \overrightarrow {c.} \overrightarrow {d.} } \right)$$
$$ = 1 + 1 + 1 + 1 + 2\left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right]$$
$$ + \left[ {\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow b } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right]$$
$$ = 4 + 2\left[ {0 + 0 + 1 \times \dfrac{1}{2} + 0 + 1 \times 1 \times \dfrac{1}{2} + 1 \times 1 \times \dfrac{1}{2}} \right]$$
$$ = 4 + 2\left[ {\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}} \right]$$
$$ = 4 + 2 \times \dfrac{3}{2}$$
$$ = 4 + 3$$
$$ = 7$$
$${\textbf{Hence, the value of }}{\mathbf{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right|^2} = 7.}$$
Question 5
1 / -0

In a plane at a given point $$\vec {A}$$ exist is vertically upward and $$\vec {B}$$ at point in north direction then find the direction at $$\vec {A}\times \vec {B}$$

A
cannot define

B
in eastern direction

C
west

D
vertically downward

Solution
Question 6
1 / -0

Let $$\hat{a}, \hat{b}$$ and $$\hat{c}$$ be three unit vectors such that $$\hat{a}=\hat{b}+(\hat{b}\times \hat{c})$$, then the possible value(s) of $${|\hat{a}+\hat{b}+\hat{c}|}^{2}$$ can be:

A
$$1$$

B
$$4$$

C
$$16$$

D
$$9$$

Solution
Question 7
1 / -0

If $$D\vec A = \vec a,$$ $$A\vec B = \vec b$$ and $$C\vec B = k\vec a$$ where $$k < 0$$ and X, Y are the mid-points of DB & DC respectively, such that $$\left| {\vec a} \right| = 17\& \left| {X\vec Y} \right| = 4$$, then k equal to -

A
$${8 \over {17}}$$

B
$${{13} \over {17}}$$

C
$${{25} \over {17}}$$

D
$${4 \over {17}}$$

Solution
Question 8
1 / -0

$$3\overline {OD} + \overline {DA} + \overline {DB} + \overline {DC} = $$

A
$$\overline {OA} + \overline {OB} - \overline {OC} $$

B
$$\overline {OA} + \overline {OB} - \overline {BD} $$

C
$$\overline {OA} + \overline {OB} + \overline {OC} $$

D
None of these

Solution
Question 9
1 / -0

Forces $$3 \vec { OA } $$, $$5 \vec { OB } $$ act along OA and OB. If their resultant passes through C on AB, then :

A
C is mid-point of AB

B
C divides AB in the ratio $$2:1$$

C
$$3AC=5CB$$

D
$$2AC=3CB$$

Solution

Draw $$ON$$ perpendicular to $$AB$$
Let $$\hat{i}$$ be the unit vector along $$ON$$
The resultant force $$\overrightarrow{R}=3\overrightarrow{A}+5\overrightarrow{B}$$ ......$$\left(1\right)$$
The angles between $$\hat{i}$$ and the forces $$\overrightarrow{R}$$, $$3\overrightarrow{A}$$ and $$5\overrightarrow{B}$$ are $$\angle{CON},\angle{AON}$$ and $$\angle{BON}$$ respectively.
$$\overrightarrow{R}.\hat{i}=3\overrightarrow{A}.\hat{i}+5\overrightarrow{B}.\hat{i}$$
$$\overrightarrow{R}.1.\cos{\angle{CON}}=3.\overrightarrow{OA}.1.\cos{\angle{AON}}+5\overrightarrow{OB}.1.\cos{\angle{BON}}$$
$$\overrightarrow{R}.\dfrac{ON}{OC}=3\overrightarrow{OA}\times \dfrac{ON}{OA}+5\overrightarrow{OB}\times \dfrac{ON}{OB}$$
$$\Rightarrow \dfrac{\overrightarrow{R}}{OC}=3+5$$
$$\Rightarrow \overrightarrow{R}=8\overrightarrow{OC}$$
We know that $$\overrightarrow{OA}=\overrightarrow{OC}+\overrightarrow{CA}$$
$$\Rightarrow 3\overrightarrow{OA}=3\overrightarrow{OC}+3\overrightarrow{CA}$$ .......$$\left(1\right)$$by multiplying the complete equation by $$3$$
We know that $$\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}$$
$$\Rightarrow 5\overrightarrow{OB}=5\overrightarrow{OC}+5\overrightarrow{CB}$$ .......$$\left(2\right)$$by multiplying the complete equation by $$5$$
Eqn$$\left(1\right)+\left(2\right)$$
$$\Rightarrow 3\overrightarrow{OA}+5\overrightarrow{OB}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$$
$$\Rightarrow \overrightarrow{R}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$$
$$\Rightarrow 8\overrightarrow{OC}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$$
$$\Rightarrow \left|3\overrightarrow{CA}\right|=\left|5\overrightarrow{CB}\right|$$
$$\therefore 3AC=5CB$$
Question 10
1 / -0

The angles of a triangles whose two sides are represented by vectors $$\sqrt(\vec { a } \times \vec { b })$$ and $$\vec{b}-(\vec { a } \vec { b })\vec{a}$$ are in the ratio

A
$$1:2:3$$

B
$$1:1:2$$

C
$$1:3:5$$

D
$$1:2:7$$

Solution