Q70 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors. If the three points $\lambda \vec{a}-2 \vec{b}+c, 2 \vec{a}+\lambda \vec{b}-2 \vec{c}, 4 \dot{a}+7 \vec{b}-8 \vec{c}$ are collinear, then $\lambda=$

  • A
    -1
  • B
    -2
  • C
    2
  • Correct answer
    D
    1

Solution

We know that if $\vec{a}, \vec{b}, \vec{c}$ are collinear then $\begin{aligned} & (\vec{a}-\vec{b})=k(\vec{a}-\vec{c}) \\ & \therefore(\lambda-2) \vec{a}+(\lambda-2) \vec{b}+3 \vec{c}=k[(\lambda-4) \vec{a}-9 \vec{b}+9 \vec{c}] \\ & \Rightarrow(\lambda-2) \vec{a}+(\lambda-2) \vec{b}+3 \vec{c}=k(\lambda-4) \vec{a}-9 k \vec{b}+9 k \vec{c} \end{aligned}$
On comparing both sides, we get $3=9 k \Rightarrow k=\frac{1}{3}$ $\begin{aligned} & \lambda-2=k(\lambda-4) \Rightarrow 3 \lambda-6=\lambda-4 \Rightarrow 2 \lambda=6-4 \Rightarrow \lambda=1 \\ & \text { And }(\lambda-2)=k(\lambda-4) \Rightarrow(1-2)=\frac{1}{3}(1-4) \\ & \Rightarrow-1=-1 \text { satisfy. } \therefore \lambda=1 \end{aligned}$

Hence, the correct answer is option: 'D'.

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