CBSE Maths › Class 12 PYQs › Vector Algebra

Vector Algebra: 2 marks Questions (CBSE Class 12)

13 different 2 marks questions on Vector Algebra from CBSE Class 12 Maths board exams 2026, newest first.

Q232 marksVery Short AnswerVector Algebra CBSE 2026 · 65/1/1

Find the vector of magnitude 14 in the direction of $QP$ , where P and Q are the points (1, 3, 2) and (-1, 0, 8) respectively.

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Answer:

4 i^+ 6 j^- 12 k^

$QP = (1 + 1) i^+ (3 - 0) j^+ (2 - 8) k^= 2 i^+ 3 j^- 6 k^$ .
$∣ QP ∣ = 4 + 9 + 36 = 7$ .
Required vector $= 14 \times \frac{2 i ^ + 3 j ^ - 6 k ^}{7} = 4 i^+ 6 j^- 12 k^$ .

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/1/1

Vectors $a = 3 i^- 2 j^+ 2 k^$ and $b = i^+ 2 k^$ represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths.

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Answer: Diagonals

4 i^- 2 j^+ 4 k^

(length 6) and

2 i^- 2 j^

(length

22

)

Diagonals are $a + b$ and $a - b$ .
$a + b = 4 i^- 2 j^+ 4 k^$ , length $16 + 4 + 16 = 6$ .
$a - b = 2 i^- 2 j^+ 0 k^$ , length $4 + 4 = 22$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q212 marksVery Short AnswerVector Algebra CBSE 2026 · 65/1/2

A vector $a$ of magnitude 14 has direction ratios $⟨ 2, 3, - 6 ⟩$ . Find the projection of the vector $a$ on $i^$ .

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Answer: 4

Direction ratios 2, 3, $- 6$ have magnitude $4 + 9 + 36 = 7$ , so direction cosines are $\frac{2}{7}, \frac{3}{7}, - \frac{6}{7}$ .
$a = 14 (\frac{2}{7} i^+ \frac{3}{7} j^- \frac{6}{7} k^) = 4 i^+ 6 j^- 12 k^$ .
Projection of $a$ on $i^$ $= a \cdot i^= 4$ .

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/1/3

Using vectors, find the area of $△ A B C$ with vertices A(1, 2, 3), B(2, -1, 4) and C(4, 5, -1).

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Answer:

\frac{274}{2}

sq units

$A B = i^- 3 j^+ k^$ , $A C = 3 i^+ 3 j^- 4 k^$ .
$A B \times A C = (12 - 3) i^- (- 4 - 3) j^+ (3 + 9) k^= 9 i^+ 7 j^+ 12 k^$ .
$∣ A B \times A C ∣ = 81 + 49 + 144 = 274$ .
Area $= \frac{1}{2} 274$ sq units.

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/2/1

If $A B = j^+ k^$ and $A C = 3 i^- j^+ 4 k^$ represent the two vectors along the sides AB and AC of $△ A B C$ , prove that the median $A D = \frac{A B + A C}{2}$ , where D is midpoint of BC.
Hence, find the length of median AD.

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Answer: Proved;

A D = \frac{34}{2}

$A D = A B + B D = A B + \frac{1}{2} B C = A B + \frac{1}{2} (A C - A B) = \frac{A B + A C}{2}$ .
$A B + A C = 3 i^+ 0 j^+ 5 k^$ , so $A D = \frac{3}{2} i^+ \frac{5}{2} k^$ .
$∣ A D ∣ = \frac{1}{2} 9 + 25 = \frac{34}{2}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/3/1

If for two unit vectors $a$ and $b$ , $∣ a + 2 b ∣ = ∣2 a - b ∣$ , then find the angle between $a$ and $b$ .

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Answer:

\frac{π}{2}

Square both sides: $∣ a ∣^{2} + 4 a \cdot b + 4∣ b ∣^{2} = 4∣ a ∣^{2} - 4 a \cdot b + ∣ b ∣^{2}$ .
With $∣ a ∣ = ∣ b ∣ = 1$ : $5 + 4 a \cdot b = 5 - 4 a \cdot b$ , so $a \cdot b = 0$ .
The angle is $\frac{π}{2}$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/4/1

If the position vectors of three points A, B and C are $3 i^+ j^$ , $5 i^+ 6 j^- 3 k^$ and $4 j^$ respectively, then show that they form an isosceles triangle.

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Answer: Proved.

$A B = 2 i^+ 5 j^- 3 k^$ , $∣ A B ∣ = 4 + 25 + 9 = 38$ .
$B C = - 5 i^- 2 j^+ 3 k^$ , $∣ B C ∣ = 25 + 4 + 9 = 38$ .
$A C = - 3 i^+ 3 j^$ , $∣ A C ∣ = 18 = 32$ .
$A B + A C > B C$ etc., so the points form a triangle with $A B = B C$ ; it is isosceles.

Q252 marksVery Short AnswerVector Algebra CBSE 2026 · 65/4/1

Let two rods placed on the ground be represented by vectors $4 i^- j^+ 3 k^$ and $- 2 i^+ j^- 2 k^$ . Find a vector representing a flag-post of height 5 m that has to be erected perpendicular to both the rods.

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Answer:

\pm \frac{5}{3} (- i^+ 2 j^+ 2 k^)

$a \times b = i^4 - 2 j^- 1 1 k^3 - 2 = - i^+ 2 j^+ 2 k^$ .
Its magnitude is $1 + 4 + 4 = 3$ , so a unit vector perpendicular to both is $\frac{1}{3} (- i^+ 2 j^+ 2 k^)$ .
Required vector of length 5: $\pm \frac{5}{3} (- i^+ 2 j^+ 2 k^)$ (the upward one is taken for the flag-post).

Also asked in: 2026 65/4/2, 2026 65/4/3

Q25 (OR) (OR)2 marksVery Short AnswerVector Algebra CBSE 2026 · 65/4/1

A unit vector $a$ is such that it makes an angle $\frac{π}{4}$ with x-axis, $\frac{π}{3}$ with y-axis and an acute angle $θ$ with z-axis. Find $θ$ and the components of $a$ .

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Answer:

θ = \frac{π}{3}

; components

\frac{1}{2}, \frac{1}{2}, \frac{1}{2}

, i.e.

a = \frac{1}{2} i^+ \frac{1}{2} j^+ \frac{1}{2} k^

$cos^{2} \frac{π}{4} + cos^{2} \frac{π}{3} + cos^{2} θ = 1 \Rightarrow \frac{1}{2} + \frac{1}{4} + cos^{2} θ = 1$ .
$cos^{2} θ = \frac{1}{4}$ ; $θ$ acute so $cos θ = \frac{1}{2}$ , $θ = \frac{π}{3}$ .
For a unit vector the components are the direction cosines: $a = \frac{1}{2} i^+ \frac{1}{2} j^+ \frac{1}{2} k^$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/4/2

If $(2 i^+ 3 j^+ k^)$ and $(2 i^+ j^- k^)$ represent the sides $A B$ and $A C$ respectively of $△ A B C$ , find the vector representing the median through A.

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Answer:

2 i^+ 2 j^

Let D be the mid-point of BC. Then $A D = \frac{1}{2} (A B + A C)$ .
$A B + A C = 4 i^+ 4 j^$ .
Median $A D = 2 i^+ 2 j^$ .

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/4/3

If in a parallelogram PQRS, $P Q = 2 i^+ 3 j^+ k^$ and $QR = i^- 2 j^+ k^$ , then find the unit vectors parallel to the diagonals $P R$ and $S Q$ .

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Answer:

\frac{1}{14} (3 i^+ j^+ 2 k^)

and

\frac{1}{26} (i^+ 5 j^)

$P R = P Q + QR = 3 i^+ j^+ 2 k^$ , $∣ P R ∣ = 14$ .
$P S = QR$ , so $S Q = P Q - P S = i^+ 5 j^$ , $∣ S Q ∣ = 26$ .
Unit vectors: $\frac{1}{14} (3 i^+ j^+ 2 k^)$ and $\frac{1}{26} (i^+ 5 j^)$ .

Q242 marksVery Short AnswerVector Algebra CBSE 2026 · 65/5/1

Three honey bees were found flying along the vectors $a = 2 i^- 3 j^+ k^$ , $b = 4 j^- 2 k^$ and $c = 3 i^+ 2 k^$ respectively. Find the value of $λ$ such that the path for $a + λ b$ is perpendicular to $c$ .

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Answer:

λ = 2

$a + λ b = 2 i^+ (4 λ - 3) j^+ (1 - 2 λ) k^$ .
Perpendicular to $c$ : $3 (2) + 0 + 2 (1 - 2 λ) = 0$ .
$8 - 4 λ = 0 \Rightarrow λ = 2$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

Q24 (OR) (OR)2 marksVery Short AnswerVector Algebra CBSE 2026 · 65/5/1

If A, B and C be three non-collinear points such that $A B = i^+ 2 j^- k^$ and $A C = 2 i^- 3 j^$ , then find the area of $△ A B C$ .

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Answer:

\frac{62}{2}

sq. units

$A B \times A C = i^12 j^2 - 3 k^- 1 0 = - 3 i^- 2 j^- 7 k^$ .
Its magnitude is $9 + 4 + 49 = 62$ .
Area $= \frac{1}{2} 62$ sq. units.

Also asked in: 2026 65/5/2, 2026 65/5/3

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