CBSE Sample Paper 2025-26 · Section A · 1 mark
Q15: (2a⋅i^)i^−(b⋅j^)j^+(c⋅k^)k^
The question
If a=3i^+2j^+4k^, b=i^+j^−3k^ and c=6i^−j^+2k^ are three given vectors, then (2a⋅i^)i^−(b⋅j^)j^+(c⋅k^)k^ is same as the vector
(A) a
(B) b+c
(C) a−b
(D) c
Exam answer
2a⋅i^=2×3=6, b⋅j^=1, c⋅k^=2.
Why?
The dot product with i^ picks out the i^-component, since i^⋅i^=1 and j^⋅i^=k^⋅i^=0:
(xi^+yj^+zk^)⋅i^=x
In the same way j^ picks out the j^-component and k^ the k^-component. So a⋅i^=3, b⋅j^=1 and c⋅k^=2.
Why?
Put the three numbers in: 6i^−(1)j^+(2)k^. The minus sign in front of the j^ term comes from the expression itself.
This has the same components as c=6i^−j^+2k^.
The others: a=3i^+2j^+4k^, b+c=7i^−k^ and a−b=2i^+j^+7k^, all different.
Why?
Add or subtract component by component:
b+c=(1+6)i^+(1−1)j^+(−3+2)k^=7i^+0j^+(−1)k^=7i^−k^
and a−b has components 3−1, 2−1 and 4−(−3):
a−b=(3−1)i^+(2−1)j^+(4+3)k^=2i^+j^+7k^
Answer: (D) c