Class 12 Mathematics Chapter 10 Vectors – Extra Questions with Answers

Extra practice questions for Class 12 Maths Chapter 10 (Vectors), beyond the textbook. These Class 12 Mathematics Chapter 10 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. Find the magnitude of a=3i+4j.
Ans: |a|=√(9+16)=√25=5.

Q2. What is the condition for two vectors to be perpendicular?
Ans: Their dot product is zero.

Q3. What is the condition for two vectors to be parallel?
Ans: Their cross product is the zero vector.

Q4. Write the formula for a unit vector along a.
Ans: a/|a|.

Q5. State the section formula for internal division in ratio m:n.
Ans: (mb+na)/(m+n).

Short Answer Questions (2–3 marks)

Q6. Find a·b if a=i+2j+3k and b=3i−2j+k.
Ans: a·b=1(3)+2(−2)+3(1)=3−4+3=2.

Q7. Find the angle between a=i+j and b=i−j.
Ans: a·b=1−1=0. Since dot product is 0, the angle is 90°.

Q8. Find a×b if a=i and b=j.
Ans: i×j=k (standard result).

Higher-Order Thinking / Application Questions

Q9. Find the area of the triangle with vertices A(1,2,3), B(2,3,1), C(3,1,2) using the vector cross product method, showing full calculation.
Ans: First find two side vectors from A: AB=B−A=(2−1,3−2,1−3)=(1,1,−2), and AC=C−A=(3−1,1−2,2−3)=(2,−1,−1). The area of the triangle is ½|AB×AC|. Compute the cross product AB×AC using the determinant: |i j k; 1 1 −2; 2 −1 −1| = i[(1)(−1)−(−2)(−1)] − j[(1)(−1)−(−2)(2)] + k[(1)(−1)−(1)(2)] = i[−1−2] − j[−1+4] + k[−1−2] = −3i−3j−3k. The magnitude is |AB×AC|=√(9+9+9)=√27=3√3. Therefore, the area of the triangle=½(3√3)=3√3/2 square units. This demonstrates how the cross product magnitude directly gives twice the triangle area, following from the geometric definition of cross product as related to the parallelogram spanned by the two vectors, of which the triangle is exactly half.

Q10. Show that the vectors a=2i−3j+4k, b=−4i+6j−8k are parallel, and explain the geometric significance of this using both the cross product and scalar multiple approaches.
Ans: Approach 1 (scalar multiple): Observe that b=−4i+6j−8k=−2(2i−3j+4k)=−2a. Since b is a scalar multiple of a (specifically −2 times a), the two vectors are parallel (though pointing in opposite directions, since the scalar is negative). Approach 2 (cross product): Compute a×b using the determinant: |i j k; 2 −3 4; −4 6 −8| = i[(−3)(−8)−(4)(6)] − j[(2)(−8)−(4)(−4)] + k[(2)(6)−(−3)(−4)] = i[24−24] − j[−16+16] + k[12−12] = 0i−0j+0k = 0 (the zero vector). Since the cross product is the zero vector, this confirms (independently of the scalar-multiple observation) that a and b are parallel — this makes sense because the cross product magnitude is |a||b|sinθ, and sinθ=0 exactly when θ=0° or 180°, i.e., when the vectors are parallel or anti-parallel. Geometrically, this means a and b lie along the same straight line through the origin, just with b twice as long as a and pointing in the opposite direction.

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.