Vector Algebra Test

Result

Vector Algebra Test - 16

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 the vectors a, b and c form the sides be the sides AB, CA, and AB respectively of a triangle ABC, then

A
$$\vec a. \vec b + \vec b. \vec c +\vec c. \vec a = 0$$

B
$$\vec a\times \vec b = \vec b\times \vec c = \vec c \times \vec a$$

C
$$\vec a. \vec b = \vec b. \vec c = \vec c. \vec a$$

D
$$\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a = 0$$

Solution

Consider the problem
For a triangle
$$\vec a + \overrightarrow {b} + \vec c = 0$$ ----- (i)
Take cross product of $$\vec b$$ on both sides

$$\vec a \times \vec a + \vec a \times \overrightarrow {b} + \vec a \times \vec c = 0$$

$$\overrightarrow {b} \times \vec a + \vec c \times \vec a = 0$$

$$\vec c \times \vec a = \vec a \times \overrightarrow {b} $$ ---- (ii)

Similarly take cross product of $$\vec b$$ both side in equation (i)

$$\begin{array}{l} \vec { a } \times \vec { b } +\vec { b } \times \overrightarrow { b } +\vec { c } \times \vec { b } =0 \\ \vec { a } \times \vec { b } =\vec { b } \times \vec { c } \, \, \, ----\, (iii) \end{array}$$
from (ii) and (iii)

$$\vec a \times \vec b = \vec b \times \vec c = \vec c \times \vec a$$

Hence option $$B$$ is correct answer.
Question 2
1 / -0

If $$\vec{a}, \vec{b}, \vec{c}$$ are three non coplanar vectors, then $$(\vec{a}+\vec{b}+\vec{c})[(\vec{a}+\vec{b}) \times (\vec{a}+\vec{c})] $$ is :

A
0

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

C
$$- [\vec{a}. \vec{b}. \vec{c}]$$

D
$$ [\vec{a}. \vec{b}. \vec{c}]$$

Solution

Since$$\vec{a}, \vec{b}, \vec{c}$$ are three non-coplanar vectors. Then say
$$(\vec{a}+\vec{b}+\vec{c}) [(\vec{a}+\vec{b}) \times (\vec{a}+\vec{c})]=\vec{x}$$
$$\Rightarrow \vec{x}=(\vec{a}+\vec{b}+\vec{c}) [(\vec{a}\times \vec{a}+ \vec{a} \times \vec{c}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c}]$$
$$=(vec{a}+\vec{b}+\vec{c})[\vec{a}\times \vec{a}+\vec{b}\times \vec{a}+\vec{b}\times \vec{c}]$$
$$[\because \vec{a}\times \vec{a}=0]$$
As $$[\vec{a} \vec{b} \vec{c}] =\vec{a}. (\vec{b} \times \vec{c})$$, so we can write
$$\vec{x}=[\vec{a}\vec{a}\vec{c}] + [\vec[{a}\vec{b} \vec{a}]+ [\vec{a} \vec{b}\vec{c}] + [\vec{b}\vec{a}\vec{c}] + [\vec{b} \vec{b} \vec{a}]+ [\vec{c} \vec{a} \vec{c}]+[\vec{c}\vec{b} \vec{c}]$$
Using $$[\vec{c}\vec[a]\vec{a}] =[\vec{b}\vec{c}\vec{a}]$$
$$=[\vec{c}\vec{a}\vec{b}]$$ & $$[\vec{a} \vec{a} \vec{c}] = [\vec{a} \vec{b} \vec{a}]$$
$$[\vec{c}\vec{a} \vec{a}]=0$$
$$\vec{x} = -[\vec{a}\vec{b} \vec{c}]$$
Question 3
1 / -0

Two or more vectors having the same initial point are

A
Coinitial vectors

B
colinear vectors

C
equal vectors

D
Cannot say

Solution

Two or more vectors having same initial points are known as co-initial vectors,

This implies that the correct option is (A)
Question 4
1 / -0

Given the points $$A(-2,3,4),B(3,2,5),C(1,-1,2),\,\&\,D(3,2,-4)$$ The projection of the vector $$\displaystyle \underset{AB}{\rightarrow}$$ on the vector $$\displaystyle \underset{CD}{\rightarrow}$$ is

A
$$\displaystyle \frac{22}{3}$$

B
$$\displaystyle -\frac{21}{4}$$

C
$$\displaystyle \frac{1}{7}$$

D
$$-47$$

Solution

The projection of the vector $$\displaystyle \underset{AB}{\rightarrow}$$ on the vector $$\displaystyle \underset{CD}{\rightarrow}$$ is $$\displaystyle \frac { \vec { AB } .\vec { CD } }{ \left| \vec { CD } \right| } =\frac { \left( 5i-j+k \right) .\left( 2i+3j-6k \right) }{ \left| \left( 2i+3j-6k \right) \right| } =\frac { 1 }{ 7 } $$
Question 5
1 / -0

The sum of two forces is $$18$$ N and resultant whose direction is at right angles to the smaller force is $$12$$N. The magnitude of the two forces are

A
$$13, 5$$

B
$$14, 4$$

C
$$12, 6$$

D
$$11, 7$$

Solution

Let the smaller force be B ,the larger force be A,resultant being C.
Given $$|A| +|B|=18 , |C|=12$$
A,B,C simply form a right angled triangle with base B vector,hypotenuse being A.
so,you have one equation with you $$A+B=18$$

$$A^2 = B^2+ C^2$$
$$=>A^2 - B^2=C^2$$
$$=>(A-B) (A+B) = 144$$
$$=>(A-B)(18)=144$$
$$=>A-B= 144/18= 8$$

Now solving both equations
$$A-B=8$$---------------------(i)
$$ A+B=18$$-------------------(ii)
adding both (i) and (ii, we get)
$$2A = 26$$
$$=> A= 13$$
Putting this value in (i)
$$13- B=8$$
$$=>B=5$$
$$|A|=13$$ and $$|B|=5$$
Question 6
1 / -0

If O is origin and C is the mid point of $$A(2, -1)$$ and $$B(-4,3)$$. Then value of OC is

A
$$i + j$$

B
$$-i + j$$

C
$$i - j$$

D
$$-i - j$$

Solution

$$\text{Since C is mid point of A(2,-1) and B(-4,3).}$$
$$\therefore C(\dfrac{2-4}{2},\dfrac{-1+3}{2})i.e C(-1,1)$$
$$\therefore OC=(-1-0)\hat i+(1-0)\hat j=-\hat i+\hat j$$
Option B is correct.
Question 7
1 / -0

Given that u is a vector of length $$2$$, v is a vector of length $$3$$ and the angle between them when placed tail to tail is $$\displaystyle 45^{\circ} $$, which option is closest to the exact value of $$\vec u\cdot\vec v$$ ?

A
$$4.5$$

B
$$6.2$$

C
$$4.2$$

D
$$5.1$$

Solution

$$u.v=|u||v|\cos (45^0)$$
$$=2.3.\cos(45^0)$$
$$=4.24$$
Option C is correct.
Question 8
1 / -0

The figure formed by the four points i + j - k, 2i + 3j, 3i + 5j -2k and k - j is

A
Rectangle

B
Trapezium

C
Parallelogram

D
None of these

Solution

$$\begin{array}{l} AB=\vec { i } +2\vec { j } +\vec { k } \\ BC=\vec { i } +2\vec { j } -2\vec { k } \\ CD=-3\vec { i } -6\vec { j } +3\vec { k } \\ DA=\vec { i } +2\vec { j } -2\vec { k } \\ \therefore BC=DA\, \, and\, \, BC\parallel DA \\ and\, \, AB\, \, is\, \, not\, \, parallel\, \, to\, \, CD \\ \therefore po{ { int } }\, made\, \, a\, \, trapezium \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer \end{array}$$
Question 9
1 / -0

ABCDEF is a regular hexagon . Let $$\displaystyle \vec { AB } =a$$ and $$\displaystyle \vec { BC } =b$$. Express the vectors $$\displaystyle \vec { AC } $$ in terms if a and b.

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

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

C
$$2\vec{a}$$

D
None of these

Solution

By the triangle law addition,
In triangle $$ABC$$
by this law sum of two vector are equal to third vector which is connecting the the tail of vector $$a$$ and head of $$b$$.
So therefore,
$$\vec {AC}=\vec a+\vec b$$
Question 10
1 / -0

The system of vectors $$i, j, k$$ is

A
Orthogonal

B
Collinear

C
Coplanar

D
None of these

Solution

$$\text{Since i,j, k represent unit vector in the direction of X,Y and Z axis respectively.}$$
$$\text{Therefore, they are orthogonal.}$$
Option A is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Vector Algebra Test - 16

Result

Vector Algebra Test - 16

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\vec a. \vec b + \vec b. \vec c +\vec c. \vec a = 0$$

$$\vec a\times \vec b = \vec b\times \vec c = \vec c \times \vec a$$

$$\vec a. \vec b = \vec b. \vec c = \vec c. \vec a$$

$$\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a = 0$$

Solution

Question 2

0

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

$$- [\vec{a}. \vec{b}. \vec{c}]$$

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

Solution

Question 3

Coinitial vectors

colinear vectors

equal vectors

Cannot say

Solution

Question 4

$$\displaystyle \frac{22}{3}$$

$$\displaystyle -\frac{21}{4}$$

$$\displaystyle \frac{1}{7}$$

$$-47$$

Solution

Question 5

$$13, 5$$

$$14, 4$$

$$12, 6$$

$$11, 7$$

Solution

Question 6

$$i + j$$

$$-i + j$$

$$i - j$$

$$-i - j$$

Solution

Question 7

$$4.5$$

$$6.2$$

$$4.2$$

$$5.1$$

Solution

Question 8

Rectangle

Trapezium

Parallelogram

None of these

Solution

Question 9

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

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

$$2\vec{a}$$

None of these

Solution

Question 10

Orthogonal

Collinear

Coplanar

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.