If (1,5,35), (7,5,5), (1, lambda, 7) and (2 l | vedclass

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(C) Let the points be $A(1, 5, 35)$,$B(7, 5, 5)$,$C(1, \lambda, 7)$,and $D(2 \lambda, 1, 2)$.The vectors are:$\vec{AB} = (7-1)\hat{i} + (5-5)\hat{j} +

Vedclass · Accepted Answer

$\frac{44}{5}$

Vedclass · Answer

(C) Let the points be $A(1, 5, 35)$,$B(7, 5, 5)$,$C(1, \lambda, 7)$,and $D(2 \lambda, 1, 2)$.The vectors are:$\vec{AB} = (7-1)\hat{i} + (5-5)\hat{j} + (5-35)\hat{k} = 6\hat{i} - 30\hat{k}$$\vec{AC} = (1-1)\hat{i} + (\lambda-5)\hat{j} + (7-35)\hat{k} = (\lambda-5)\hat{j} - 28\hat{k}$$\vec{AD} = (2\lambda-1)\hat{i} + (1-5)\hat{j} + (2-35)\hat{k} = (2\lambda-1)\hat{i} - 4\hat{j} - 33\hat{k}$Since the points are coplanar,the scalar triple product $[\vec{AB} \vec{AC} \vec{AD}] = 0$.$\begin{vmatrix} 6 & 0 & -30 \ 0 & \lambda-5 & -28 \ 2\lambda-1 & -4 & -33 \end{vmatrix} = 0$Expanding along the first row:$6[(\lambda-5)(-33) - (-28)(-4)] - 0 + (-30)[0 - (\lambda-5)(2\lambda-1)] = 0$$6[-33\lambda + 165 - 112] - 30[-(2\lambda^2 - \lambda - 10\lambda + 5)] = 0$$6[-33\lambda + 53] + 30[2\lambda^2 - 11\lambda + 5] = 0$Divide by $6$:$-33\lambda + 53 + 5(2\lambda^2 - 11\lambda + 5) = 0$$-33\lambda + 53 + 10\lambda^2 - 55\lambda + 25 = 0$$10\lambda^2 - 88\lambda + 78 = 0$$5\lambda^2 - 44\lambda + 39 = 0$The sum of the roots $\lambda_1 + \lambda_2 = -\frac{b}{a} = -\frac{-44}{5} = \frac{44}{5}$.

If $(1,5,35), (7,5,5), (1, \lambda, 7)$ and $(2 \lambda, 1, 2)$ are coplanar,then the sum of all possible values of $\lambda$ is

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