If 3 hat{i}-2 hat{j}-hat{k},2 hat{i}+3 hat{j} | vedclass

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(D) Given position vectors are:$P = 3 \hat{i} - 2 \hat{j} - \hat{k}$$Q = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$$R = -\hat{i} + \hat{j} + 2 \hat{k}$$S = 4

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$-\frac{146}{17}$

Vedclass · Answer

(D) Given position vectors are:$P = 3 \hat{i} - 2 \hat{j} - \hat{k}$$Q = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$$R = -\hat{i} + \hat{j} + 2 \hat{k}$$S = 4 \hat{i} + 5 \hat{j} + \lambda \hat{k}$Since the points $P, Q, R, S$ are coplanar,the scalar triple product of vectors $\vec{PQ}, \vec{PR}, \vec{PS}$ must be zero,i.e.,$\vec{PS} \cdot (\vec{PQ} 	imes \vec{PR}) = 0$.First,calculate the vectors:$\vec{PQ} = Q - P = (2-3)\hat{i} + (3-(-2))\hat{j} + (-4-(-1))\hat{k} = -\hat{i} + 5\hat{j} - 3\hat{k}$$\vec{PR} = R - P = (-1-3)\hat{i} + (1-(-2))\hat{j} + (2-(-1))\hat{k} = -4\hat{i} + 3\hat{j} + 3\hat{k}$$\vec{PS} = S - P = (4-3)\hat{i} + (5-(-2))\hat{j} + (\lambda-(-1))\hat{k} = \hat{i} + 7\hat{j} + (\lambda+1)\hat{k}$Now,calculate the cross product $\vec{PQ} 	imes \vec{PR}$:$\vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 5 & -3 \ -4 & 3 & 3 \end{vmatrix} = \hat{i}(15 - (-9)) - \hat{j}(-3 - 12) + \hat{k}(-3 - (-20)) = 24\hat{i} + 15\hat{j} + 17\hat{k}$Now,compute the dot product $\vec{PS} \cdot (\vec{PQ} 	imes \vec{PR}) = 0$:$(\hat{i} + 7\hat{j} + (\lambda+1)\hat{k}) \cdot (24\hat{i} + 15\hat{j} + 17\hat{k}) = 0$$1(24) + 7(15) + 17(\lambda+1) = 0$$24 + 105 + 17\lambda + 17 = 0$$146 + 17\lambda = 0$$17\lambda = -146$$\lambda = -\frac{146}{17}$

If $3 \hat{i}-2 \hat{j}-\hat{k}$,$2 \hat{i}+3 \hat{j}-4 \hat{k}$,$-\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are respectively the position vectors of four coplanar points $P, Q, R$ and $S$,then $\lambda=$

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