Let a, b and c be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be a - b + c, λa - 3b + 4c, -a + 2b - 3c and 2a - 4b + 6c respectively. If overrightarrowAB, overrightarrowAC and overrightarrowAD are coplanar, then λ is equal to

Let a, b and c be three non-zero non-coplanar vectors. Let… — JEE Main 2023 Mathematics PYQ

JEE Main 2023 · 29 Jan · Shift 1 · numerical answer

Let

\vec{a}

\vec{b}

and

\vec{c}

be three non-zero non-coplanar vectors. Let the position vectors of four points

A, B, C

and

D

\vec{a} - \vec{b} + \vec{c}

\lambda\vec{a} - 3\vec{b} + 4\vec{c}

-\vec{a} + 2\vec{b} - 3\vec{c}

and

2\vec{a} - 4\vec{b} + 6\vec{c}

respectively. If

\overrightarrow{AB}

\overrightarrow{AC}

and

\overrightarrow{AD}

are coplanar, then

\lambda

is equal to

Show answer and solution

Answer: 02

Solution

AB = (λ-1)a - 2b + 3c
AC = -2a + 3b - 4c
AD = a - 3b + 5c

For coplanarity, the determinant of coefficients must be zero:

|λ-1 -2 3|
| -2 3 -4| = 0
| 1 -3 5|

Expanding: (λ-1)(3·5 - (-4)(-3)) - (-2)((-2)·5 - (-4)·1) + 3((-2)(-3) - 3·1)

= (λ-1)(15 - 12) + 2(-10 + 4) + 3(6 - 3)

= 3(λ-1) + 2(-6) + 3(3)

= 3λ - 3 - 12 + 9 = 3λ - 6 = 0

λ = 2

[JEE Main 2023 · 29 Jan · Shift 1]

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