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All questions of Vector Algebra for JEE Exam

The area of triangle whose adjacent sides are is :
  • a)
    √70/2 sq. units
  • b)
    9√2 /2 sq. units
  • c)
    3√3 /2 sq. units
  • d)
    2√3 /2 sq. units
Correct answer is option 'A'. Can you explain this answer?

Suresh Iyer answered
Area of triangle = ½(a * b)
a = (1, 0, -2)   b = (2, 3, 1)
= i(0 + 6) + j(-4 - 1) + k(3 - 0)
= 6i - 5j + 3k
|a * b| = (36 + 25 + 9)½
|a * b| = (70)½
Area of triangle = ½(a * b)
= [(70)½]/2

A vector of magnitude 14 units, which is parallel to the vector
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

EduRev JEE answered
Given vector = i + 2j - 3k
Magnitude = √12 + 22 + (-3)2 = √14
Unit vector in direction of resultant = (i + 2j - 3k) / √14
Vector of magnitude 14​ unit in direction of resultant,
⇒ 14[ (i + 2j - 3k) / √14 ]
⇒ √14(i + 2j - 3k)

The value of  is:
  • a)
    0
  • b)
    3
  • c)
    1/3
  • d)
    1
Correct answer is option 'B'. Can you explain this answer?

Sahil Soni answered
Cross multiply in maths take place in cycle like i》j》k i×j=k j×k=i k×i=j but j×i=-k k×j=-i i×k=-j and dotmultiply takes place as i.i=1j.j=1K.K=1BUTI.J=0J.K=0k.i=0so the correct answer is b

The points with position vectors  are collinear vectors, Value of a =​
  • a)
    -20
  • b)
    20
  • c)
    -40
  • d)
    40
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
Position vector A = 60i+3j
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear,  angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of  the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.

The unit vector in the direction of , where A and B are the points (2, – 3, 7) and (1, 3, – 4) is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Sushil Kumar answered
Given, Point A (2,-3,7)
Point B (1,3,-4)
Let vector in the direction of AB be C.
∴ C = B - A
⇒ (1,3,-4) - (2,-3,7)
⇒ ( 1-2 , 3+3 , -4-7 )
⇒ (-1,6,-11)
⇒ -1i + 6j -11k
Magnitude of vector C
|C| = √(-1)2 + 62 + (-11)2
⇒ √1+36+121
⇒ √158
Unit vector = (Vector)/(Magnitude of vector)
Unit vector C = (C vector)/(Magnitude of C vector)  = (-1i + 6j -11k)/√158

If  and , then the value of scalars x and y are:
  • a)
    x = 1 and y = -2
  • b)
    x = -2 and y = 1
  • c)
    x = 2 and y = -1
  • d)
    x = 2 and y = 1
Correct answer is option 'C'. Can you explain this answer?

Sushil Kumar answered
Given, a = i + 2j
b = -2i + j
c = 4i +3j
Also, c = xa +yb
Now putting the values in above equation,
4i + 3j  = x(i + 2j) + y(-2i +j)
⇒ xi + 2xj - 2yi + yj
⇒ (x-2y)i + (2x+y)j
We get,
x - 2y = 4
2x + y = 3
After solving,            
x = 2
y = -1

In the triangle ABC, which statement is not true?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

sanju sura answered
Ab + bc > ac ...it is property of a triangle .... that sum of two sides is greater than the third side ... so I think it is no possible that ab +bc -ca =0...but I'm not understanding the logic behind the question.......😊

If a and b are the position vectors of two points A and B and C is a point on AB produced such that AC = 3AB, then position vector of C will be​
  • a)
    3b – 2a
  • b)
    3a – b
  • c)
    3a – 2b
  • d)
    3b – a
Correct answer is option 'A'. Can you explain this answer?

Anuj Datta answered
Given information:

- Position vector of point A: a
- Position vector of point B: b
- Point C is on line AB produced such that AC = 3AB

To find:

- Position vector of point C

Solution:

Step 1: Find the vector AB

The vector AB is given by:

AB = b - a

Step 2: Find the vector AC

Since AC = 3AB, we can write:

AC = 3(b - a)

Step 3: Find the position vector of C

The position vector of C can be obtained by adding the vector AC to the position vector of A:

C = A + AC

Substituting the values of A and AC, we get:

C = a + 3(b - a)

Simplifying, we get:

C = 3b - 2a

Therefore, the position vector of point C is 3b - 2a.

Answer: Option A (3b - 2a)

If the magnitude of the position vector is 7, the value of x is:​
  • a)
    ±1
  • b)
    ±5
  • c)
    ±3
  • d)
    ±2
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered

|a| = (x2 + 22 + (2x)2)1/2
7 = (x2 + 22 + (2x)2)1/2
⇒ 49 = x2 + 22 + 4x2
⇒ 49 = 4 + 5x2
⇒ 5x2 = 45
⇒ x2 = 9
x = ±3

If  are two vectors, such that , then = ……​
  • a)
    3
  • b)
    √7
  • c)
    √5
  • d)
    √3
Correct answer is option 'D'. Can you explain this answer?

Neha Sharma answered
 |a - b|2 = |a|2 + |b|2 - 2|a||b|
|a - b|2  = (3)2 + (2)2 - 2(5)
|a - b|2  = 9 + 4 - 10
|a - b|2  = 3 
|a - b|   = (3)½.

What is the additive identity of a vector?​
  • a)
    zero vector
  • b)
    Negative of the vector
  • c)
    unit vector
  • d)
    The vector itself
Correct answer is option 'A'. Can you explain this answer?

Nandini Iyer answered
In the Additive Identity of vectors, the additive identity is zero vector 0.
For any vector V additive identity is defined as,
0 + V = V and V + 0 = V

If  are unit vectors along X-axis, Y-axis and Z-axis respectively, then
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

As per the rules i×i=0 and i×k=-j and also j=-1 so -1×-1=1.So (A)is correct answer because i×j = k is correct.

For any two vectors a and b​, we always have
  • a)
    |a – b| ≥ |a| – |b|
  • b)
    |a + b| ≤ |a| + |b|
  • c)
    |a + b| ≤ |a| – |b|
  • d)
    |a – b| = |a + b|
Correct answer is option 'B'. Can you explain this answer?

Nandini Iyer answered
|a + b|2 = |a|2 + |b|2 + 2|a||b|.cosθ
|a|2 + |b|2 = |a|2 + |b|2 + 2|a| + |b|  ∵ −1 ⩽ cosθ ⩽ 1
⇒ 2|a||b|.cosθ ⩽ 2|a||b|
So, |a + b|2 ⩽ (|a| + |b|
)2

⇒ |a + b| ≤ |a| + |b|
This is also known as Triangle Inequality of vectors.

The vector joining the points A(2, – 3, 1) and B(1, – 2, – 5) directed from B to A is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Siddhant Kumar answered
The vector goes from B to A means that its initial coordinates is at B and final at A.
so the vector BA will be [ (2-1)i + (-3+2) j + (1+5)k ] = i-j+6k.

The value of k such that  lies in the plane 2x – 4y + z = 7, is
  • a)
    7
  • b)
    –7
  • c)
    no real value
  • d)
    4
Correct answer is option 'A'. Can you explain this answer?

Gaurav Kumar answered
As the line  lies in th e plan e 2x - 4 y +z= 7, the point (4, 2, k) through which line passes must also lie on the given plane and hence 2 × 4 – 4 × 2 + k = 7  ⇒ k = 7

If  , then
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Lavanya Menon answered
a = 2i + 3j - 6k
|a| = √4+9+36 = √49 = 7
b = 6i - 2j + 3k
|b| = √36+4+9 = √49 = 7
|a| = |b|
Hence, option A is correct.

A line with positive direction cosines passes through the point P(2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals
  • a)
    1
  • b)
    √2
  • c)
    √3
  • d)
    2
Correct answer is option 'C'. Can you explain this answer?

Samarth Saha answered
The line has +ve and equal direction cosines, these are   or direction ratios are 1, 1, 1. Also thelines passes through P (2, – 1, 2).
∴ Equation  of line is
 be a point on this line where it meets the plane 2 x + y + z = 9
Then Q must satisfy the eqn of plane
∴ Q has coordintes (3, 0, 3)

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