Vector Algebra Test

Result

Vector Algebra Test - 32

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

If $$\vec a$$ is a non-zero vector of modulus $$a$$ and $$m$$ is a non-zero scalar, then $$m \vec a$$ is a unit vector if

A
$$m = \pm 1$$

B
$$a = |m|$$

C
$$a = \dfrac{1}{|m|}$$

D
$$a = \dfrac{1}{m}$$

Solution

$$m\vec a$$ is a unit vector if and only if
$$|m \vec a| = 1 $$
$$\Rightarrow |m||\vec a| = 1$$
$$ \Rightarrow |m| a = 1$$
$$ \Rightarrow \displaystyle \dfrac{1}{|m|}$$
Question 2
1 / -0

If vector $$\vec{a} = 2\hat i - 3\hat j + 6\hat k$$ and vector $$\vec{b} = - 2\hat i + 2\hat j - \hat k$$, then ratio of Projection of $$\vec a$$ on vector $$\vec b$$ to Projection of $$\vec b$$ on $$\vec a$$ is equal to

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

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

C
$$3$$

D
$$7$$

Solution

From the condition given we get

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

Now we know that $$a.b=b.a$$ ...(by property of scalar/dot product of vectors).

Hence the above expression simplifies as
$$\dfrac{|a|}{|b|}$$

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

The projection of the line segment joining the points A(-1, 0, 3) and B(2, 5, 1) on the line whose direction ratios are proportional to 6, 2, 3, is

A
$$\dfrac{10}{7}$$

B
$$\dfrac{22}{7}$$

C
$$\dfrac{18}{7}$$

D
none of these

Solution

Given points of line segment
$$A(-1,0,3)\ and\ B(2,5,1)$$
direction ratio of line be $$R(2+1,5-0,1-3)=R(3,5,-2)$$
$$a_{1}=3,b_{1}=5,c_{1}=-2$$
$$a_{2}=6,b_{2}=2,c_{2}=3$$

projection of lines $$l_{1}\ on\ l_{2} =\dfrac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}}$$
$$\Rightarrow projection=\dfrac{3\times6+5\times2+3\times(-2)}{\sqrt{6^2+2^2+3^2}}$$
$$\Rightarrow projection=\dfrac{18+10-6}{\sqrt{36+4+9}}$$
$$\Rightarrow projection=\dfrac{22}{\sqrt{49}}$$
$$\Rightarrow projection=\dfrac{22}{7}$$
Question 4
1 / -0

Let $$\vec {p}$$ is the position vector of the orthocentre & $$\vec {g}$$ is the position vector of the centroid of the triangle $$ABC$$ where circumcentre is the origin. If $$\vec {p}= K\vec{g},$$ then $$K=$$

A
$$3$$

B
$$2$$

C
$$\displaystyle\dfrac{1}{3}$$

D
$$\displaystyle\dfrac{2}{3}$$

Solution

$$\vec p = \vec a + \vec b + \vec c$$ as per the orthocentre definition.
Also, centroid $$\vec g = \dfrac{\vec a + \vec b + \vec c}{3}$$ where, $$\vec a, \vec b, \vec c$$ are the position vectors of points $$A, B, C$$ of the triangle respectively.
Question 5
1 / -0

If $$\vec{a}, \vec{b} $$ and $$ \vec {c} $$ are three non- coplanar vectors, then the length of projection of vector $$\vec{a} $$ in the plane of the vectors $$\vec{b}$$ and $$\vec{c}$$ may be given by

A
$$\dfrac{| \vec{a}.(\vec{b}\times \vec{c})|}{|\vec{b} \times \vec{c}|}$$

B
$$\dfrac{| \vec{a}\times(\vec{b}\times \vec{c})|}{|\vec{b} \times \vec{c}|}$$

C
$$\dfrac{[\vec{a} \vec{b} \vec{c}]}{(\vec{b}. \vec{c})}$$

D
none of these

Solution

if the planes contains $$\vec{b}\ and\ \vec{c}$$ then normal vector of plane
$$\vec{b}\times\vec{c}$$
projection formula
projection of $$\vec{a}$$ on plane$$=\vec{a}\times\hat{(\vec{b}\times\vec{c})}$$
$$length=\dfrac{\left |\vec{a}\times(\vec{b}\times\vec{c})\right |}{\left |\vec{b}\times\vec{c}\right |}$$
Question 6
1 / -0

Five coplanar forces (each of magnitude $$20N$$) are acting on a body. The angle between two neighboring forces have the same value. The resultant of these forces is necessarily equal to

A
$$20N$$

B
$$20\sqrt{2}N$$

C
Zero

D
none of these

Solution

As the angle between any two neighbouring forces have the same value which is $$\dfrac{360^{\circ}}{5}=72^{\circ}$$
and all the forces are of equal magnitude.
Hence, their resultant is zero.
Question 7
1 / -0

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

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

B
$$6$$ units

C
$$- 6$$ units

D
$$3$$ units

Solution

Projection of $$x\hat i -\hat j + \hat k$$ on $$2\hat i - \hat j + 5\hat k$$
$$= \displaystyle \dfrac{(x\hat i - \hat j + \hat k) (2\hat i - \hat j + 5\hat k)}{\sqrt{4 + 1 + 25}} = \dfrac{2x + 1 + 5}{\sqrt{30}}$$
But given,
$$\displaystyle \dfrac{2x + 6}{\sqrt{30}} = \dfrac{1}{\sqrt{30}}$$
$$\Rightarrow 2x + 6 = 1$$
$$\displaystyle \Rightarrow x = \dfrac{-5}{2}$$
Question 8
1 / -0

A man starts from $$O$$ moves $$500m$$ turns by $$60 ^\circ$$ and moves $$500m$$ again turns by $$60 ^\circ$$ and moves $$500m$$ and so on. Find the displacement after $$(i)$$ 5th turn , $$(ii)$$ 3rd turn.

A
-$$500m, 1000m$$

B
$$500m, 500 \sqrt {3} m$$

C
$$1000m, 500 \sqrt {3} m$$

D
none of these

Solution

(i) Fifth turn is at E and this displacement will be given by vector $$\vec{OE}$$ which is clear from the given fig that it is- $$500m$$.
(ii) Third turn is at C and this will be given by vector $$\vec{OC}$$ which is equal to $$\sqrt{AC^2+AB^2}=\sqrt{(500\sqrt{3})^2+500^2}=1000m$$ where $$AC=2AB \ sin60^o$$
Question 9
1 / -0

The projection of the vector $$\hat i - 2\hat j + \hat k$$ on the vector $$4\hat i - 4\hat j + 7\hat k$$ is

A
$$\displaystyle \dfrac{5\sqrt{6}}{10}$$

B
$$\displaystyle \dfrac{19}{9}$$

C
$$\displaystyle \dfrac{9}{19}$$

D
$$\displaystyle \dfrac{\sqrt{6}}{19}$$

Solution

Projection of $$\vec a$$ on $$\vec b$$
$$= \displaystyle |a| \cos \theta = |a| \dfrac{a \cdot b}{|a||a|} = \dfrac{a \cdot b}{|b|}$$

$$= \displaystyle \dfrac{4 + 8 + 7}{\sqrt{16 + 16 + 49}} $$

$$= \dfrac{19}{\sqrt{81}} $$

$$= \dfrac{19}{9}$$
Question 10
1 / -0

If the vector product of a constant vector $$\displaystyle \vec{OA}$$ with a variable vector $$\displaystyle \vec{OB}$$ in a fixed plane $$OAB$$ be a constant vector, then locus of $$B$$ is :

A
A straight line perpendicular to $$\displaystyle \vec{OA}$$

B
A circle with centre O radius equal to $$\displaystyle \left | \vec{OA} \right |$$

C
A straight line parallel to $$\displaystyle \vec{OA}$$

D
None of these

Solution

Vector product of a vector with some vector in the plane, leads a constant vector; implies that the answer can only be $$0$$.
Thus, we will have a vector parallel to the original vector so that the angle between the two becomes $$0$$ and the answer comes out to be $$0$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 32

Result

Vector Algebra Test - 32

Score

-

Rank

-

SHARING IS CARING

Question 1

$$m = \pm 1$$

$$a = |m|$$

$$a = \dfrac{1}{|m|}$$

$$a = \dfrac{1}{m}$$

Solution

Question 2

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

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

$$3$$

$$7$$

Solution

Question 3

$$\dfrac{10}{7}$$

$$\dfrac{22}{7}$$

$$\dfrac{18}{7}$$

none of these

Solution

Question 4

$$3$$

$$2$$

$$\displaystyle\dfrac{1}{3}$$

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

Solution

Question 5

$$\dfrac{| \vec{a}.(\vec{b}\times \vec{c})|}{|\vec{b} \times \vec{c}|}$$

$$\dfrac{| \vec{a}\times(\vec{b}\times \vec{c})|}{|\vec{b} \times \vec{c}|}$$

$$\dfrac{[\vec{a} \vec{b} \vec{c}]}{(\vec{b}. \vec{c})}$$

none of these

Solution

Question 6

$$20N$$

$$20\sqrt{2}N$$

Zero

none of these

Solution

Question 7

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

$$6$$ units

$$- 6$$ units

$$3$$ units

Solution

Question 8

-$$500m, 1000m$$

$$500m, 500 \sqrt {3} m$$

$$1000m, 500 \sqrt {3} m$$

none of these

Solution

Question 9

$$\displaystyle \dfrac{5\sqrt{6}}{10}$$

$$\displaystyle \dfrac{19}{9}$$

$$\displaystyle \dfrac{9}{19}$$

$$\displaystyle \dfrac{\sqrt{6}}{19}$$

Solution

Question 10

A straight line perpendicular to $$\displaystyle \vec{OA}$$

A circle with centre O radius equal to $$\displaystyle \left | \vec{OA} \right |$$

A straight line parallel to $$\displaystyle \vec{OA}$$

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.