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(D) Let the position vectors of the four points be $\vec{a} = 3\hat{i}-2\hat{j}-\hat{k}$,$\vec{b} = 2\hat{i}+3\hat{j}-4\hat{k}$,$\vec{c} = -\hat{i}+\h

Vedclass · Accepted Answer

$\frac{146}{17}$

Vedclass · Answer

(D) Let the position vectors of the four points be $\vec{a} = 3\hat{i}-2\hat{j}-\hat{k}$,$\vec{b} = 2\hat{i}+3\hat{j}-4\hat{k}$,$\vec{c} = -\hat{i}+\hat{j}+2\hat{k}$,and $\vec{d} = 4\hat{i}+5\hat{j}+\lambda\hat{k}$.For the four points to be coplanar,the scalar triple product of vectors $\vec{AB}$,$\vec{AC}$,and $\vec{AD}$ must be zero.First,calculate the vectors:$\vec{AB} = \vec{b} - \vec{a} = (2-3)\hat{i} + (3-(-2))\hat{j} + (-4-(-1))\hat{k} = -\hat{i} + 5\hat{j} - 3\hat{k}$$\vec{AC} = \vec{c} - \vec{a} = (-1-3)\hat{i} + (1-(-2))\hat{j} + (2-(-1))\hat{k} = -4\hat{i} + 3\hat{j} + 3\hat{k}$$\vec{AD} = \vec{d} - \vec{a} = (4-3)\hat{i} + (5-(-2))\hat{j} + (\lambda-(-1))\hat{k} = \hat{i} + 7\hat{j} + (\lambda+1)\hat{k}$Now,set the determinant of these vectors to zero:$\begin{vmatrix} -1 & 5 & -3 \ -4 & 3 & 3 \ 1 & 7 & \lambda+1 \end{vmatrix} = 0$Expanding along the first row:$-1(3(\lambda+1) - 21) - 5(-4(\lambda+1) - 3) - 3(-28 - 3) = 0$$-1(3\lambda + 3 - 21) - 5(-4\lambda - 4 - 3) - 3(-31) = 0$$-1(3\lambda - 18) - 5(-4\lambda - 7) + 93 = 0$$-3\lambda + 18 + 20\lambda + 35 + 93 = 0$$17\lambda + 146 = 0$$17\lambda = -146$$\lambda = -\frac{146}{17}$

If the points having the position vectors $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}+3 \hat{j}-4 \hat{k}, -\hat{i}+\hat{j}+2 \hat{k}$ and $4 \hat{i}+5 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

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