Let a→,b→ and c→ be three non-zero non-coplanar vectors. Let the position vector of four points A,B, C and D be a→−b→+c→,λa→−3b→+4c¯,−a→+2b→−3c→ and 2a→−4b→+6c→ respectively. If AB→,AC→ and AD→ are coplanar, then λ is equal to ____________

AB→=B→−A→=(λ−1)a→−2b→+3c→ AC→=C→−A→=−2a→+3b→−4c→ AD→=a→−3b→+5c→ Since, AB→,AC→ and AD→ are co-planar So, |λ−1−23−23−41−35|=0 ⇒(λ−1)(15−12)+2(−10+4)+3(6−3)=0

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Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vector of four points A,B, C and D be $\vec{a} - \vec{b} + \vec{c}, λ \vec{a} - 3 \vec{b} + 4 \bar{c}, - \vec{a} + 2 \vec{b} - 3 \vec{c}$ and $2 \vec{a} - 4 \vec{b} + 6 \vec{c}$ respectively. If $\vec{A B}, \vec{A C}$ and $\vec{A D}$ are coplanar, then $λ$ is equal to ____________

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Detailed Solution

$\vec{A B} = \vec{B} - \vec{A} = (λ - 1) \vec{a} - 2 \vec{b} + 3 \vec{c} \vec{A C} = \vec{C} - \vec{A} = - 2 \vec{a} + 3 \vec{b} - 4 \vec{c} \vec{A D} = \vec{a} - 3 \vec{b} + 5 \vec{c} S i n c e, \vec{A B}, \vec{A C} a n d \vec{A D} a r e c o - p l a n a r S o, | \begin{matrix} λ - 1 & - 2 & 3 \\ - 2 & 3 & - 4 \\ 1 & - 3 & 5 \end{matrix} | = 0 \Rightarrow (λ - 1) (15 - 12) + 2 (- 10 + 4) + 3 (6 - 3) = 0$