If the points with position vectors hat{i}-ha | vedclass

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(C) Let the position vectors of the points be $\vec{A} = \hat{i} - \hat{j} + \hat{k}$, $\vec{B} = 2\hat{i} - \hat{k}$, $\vec{C} = \hat{j} + 2\hat{k}$,

Vedclass · Accepted Answer

$7$

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(C) Let the position vectors of the points be $\vec{A} = \hat{i} - \hat{j} + \hat{k}$, $\vec{B} = 2\hat{i} - \hat{k}$, $\vec{C} = \hat{j} + 2\hat{k}$, and $\vec{D} = \hat{i} + \hat{j} + \lambda\hat{k}$.For the points to be coplanar, the scalar triple product of the vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$ must be zero.$\vec{AB} = \vec{B} - \vec{A} = (2-1)\hat{i} + (0-(-1))\hat{j} + (-1-1)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$.$\vec{AC} = \vec{C} - \vec{A} = (0-1)\hat{i} + (1-(-1))\hat{j} + (2-1)\hat{k} = -\hat{i} + 2\hat{j} + \hat{k}$.$\vec{AD} = \vec{D} - \vec{A} = (1-1)\hat{i} + (1-(-1))\hat{j} + (\lambda-1)\hat{k} = 2\hat{j} + (\lambda-1)\hat{k}$.The condition for coplanarity is $\begin{vmatrix} 1 & 1 & -2 \ -1 & 2 & 1 \ 0 & 2 & \lambda-1 \end{vmatrix} = 0$.Expanding the determinant: $1(2(\lambda-1) - 2) - 1(-1(\lambda-1) - 0) - 2(-2 - 0) = 0$.$1(2\lambda - 2 - 2) + 1(\lambda - 1) + 4 = 0$.$2\lambda - 4 + \lambda - 1 + 4 = 0$.$3\lambda - 1 = 0 \Rightarrow \lambda = \frac{1}{3}$.Now, the vector is $6\lambda\hat{i} - 3\hat{j} + 6\hat{k} = 6(\frac{1}{3})\hat{i} - 3\hat{j} + 6\hat{k} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.The magnitude is $\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

If the points with position vectors $\hat{i}-\hat{j}+\hat{k}$, $2 \hat{i}-\hat{k}$, $\hat{j}+2 \hat{k}$ and $\hat{i}+\hat{j}+\lambda \hat{k}$ are coplanar, then the magnitude of the vector $6 \lambda \hat{i}-3 \hat{j}+6 \hat{k}$ is

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