If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\overline{\mathrm{a}}+2 \overline{\mathrm{~b}}+3 \overline{\mathrm{c}}, \lambda \overline{\mathrm{b}}+4 \overline{\mathrm{c}}$ and $(2 \lambda-1) \overline{\mathrm{c}}$ are non-coplanar for

(C) all except two values of $\lambda$

(B) all except one value of $\lambda$

JEE Main · 2004 · 24 Apr 2004 · Offline · Vector Algebra

\overline{a}, \overline{b}, \overline{c}

are non-coplanar vectors and

λ

is a real number, then the vectors

\overline{a} + 2 \overline{b} + 3 \overline{c}, λ \overline{b} + 4 \overline{c}

and

(2 λ - 1) \overline{c}

are non-coplanar for

(A)all values of $λ$
(B)all except one value of $λ$
(C)all except two values of $λ$ ✓ Correct
(D)no value of $λ$

Answer: (C)

Solution

Because

a, b, c

are non-coplanar, the three given vectors are non-coplanar exactly when the determinant of their coefficients is non-zero. That determinant is triangular,

100 2 λ 0 34 2 λ - 1 = λ (2 λ - 1)

, which vanishes only at

λ = 0

and

λ = \frac{1}{2}

. So the vectors are non-coplanar for every value of

λ

except those two.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

Practise more: all JEE Main PYQs on Vector Algebra →

JEE Main · 2004 · 24 Apr 2004 · Offline · Vector Algebra

Solution

More JEE Main questions on Vector Algebra