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Mathematics Question on Vector Algebra

If the four points with position vectors −2i^+j^+k,i^+j^+k^,j^−k^,-2\hat{i}+\hat{j}+k, \hat{i} +\hat{j}+\hat{k}, \hat{j}-\hat{k}, and λj^+k^\lambda\hat{j}+\hat{k} are coplanar, then λ=\lambda=

A

1

B

2

C

-1

D

0

Answer

1

Explanation

Solution

If four points (x1,y1,z1),(x2,y2,z2),(x3,y3,z3)\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right),\left(x_{3}, y_{3}, z_{3}\right) and
(x4,y4,z4)\left(x_{4}, y_{4}, z_{4}\right) are coplanar, then
∣x2−x1y2−y1z2−z1 x3−x1y3−y1z3−z1 x4−x1y4−y1z4−z1∣=0\begin{vmatrix}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\\ x_{4}-x_{1} & y_{4}-y_{1} & z_{4}-z_{1}\end{vmatrix}=0
Now, ∣300 20−2 2λ−10∣=0\begin{vmatrix}3 & 0 & 0 \\\ 2 & 0 & -2 \\\ 2 & \lambda-1 & 0\end{vmatrix}=0
⇒3(0+2λ−2)=0\Rightarrow 3(0+2 \lambda-2)=0
⇒λ=1\Rightarrow \lambda=1