Question

If p = 2a – 3b, q = a – 2b + c, r = –3a + b + 2c, where a, b, c are non-zero are non-coplanar vectors, then the vector –2a + 3b – c is equal to

(a) p – 4q

(b) -(7/5)p + (1/5)r 

(c) 2p – 3q – r

(d) – 4p – 2r

Answer 1

Answer is (b) – (7/5)q + (1/5)r

Explanation:

Since a, b and c are non-zero and non-coplanar vectors, so any vector can be expressed as a linear combination of these three vectors, i.e.

–2a + 3b – c = lp + mq + nr, where l, m, n ∈ R

–2a + 3b – c = l(2a – 3b) + m(a – 2b + c) + n(– 3a + b + 2c)

= (2l + m – 3n)a +(– 3l – 2m + n)b + (m + 2n) c

Comparing, we get (2l + m – 3n) = –2,

(– 3l – 2m + n) = 3 and (m + 2n) = –1

Solving, we get

l = 0, m = – (7/5), n = (1/5) 

Therefore, –2a + 3b – c = – (7/5)q + (1/5)r