Question

# Find the value of $$\lambda$$ for which the four points with position vectors $$6\hat{i } - 7\hat{j}, 16\hat{i} -19\hat{j}-4\hat{k}, \lambda \hat{j}- 6\hat{k}$$ and $$2\hat{i}-5\hat{j} +10\hat{k}$$ are coplanar.

Solution

## Let $$\vec { A } =6i-7j,\vec { B } =16-19j-4k=0,\vec { C } =\lambda j-6k$$ and $$\vec { D } =2i-5j+10k$$$$\vec { AB } =10i-12j-4k\\ \vec { AC } =-6i+(\lambda +7)j-6k\\ \vec { AD } =-4i+2j+10k$$Now the vectors are coplanar$$\Rightarrow \vec { AB } .(\vec { AC } \times \vec { AD } )=0\\ \Rightarrow \left| \begin{matrix} 10 & -12 & -4 \\ -6 & \lambda +7 & -6 \\ -4 & 2 & 10 \end{matrix} \right| =0\\ \Rightarrow 10(10\lambda +70+12)+12(-60-24)-4(-12+4\lambda +28)=0\\ \Rightarrow 100\lambda +820-1008-64-16\lambda =0\\ \Rightarrow 84\lambda =252\\ \Rightarrow \lambda =3$$Maths

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