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(C) Let the position vectors of the vertices be $\vec{a} = \hat{i}-6 \hat{j}+10 \hat{k}$,$\vec{b} = -\hat{i}-3 \hat{j}+7 \hat{k}$,$\vec{c} = 5 \hat{i}

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let the position vectors of the vertices be $\vec{a} = \hat{i}-6 \hat{j}+10 \hat{k}$,$\vec{b} = -\hat{i}-3 \hat{j}+7 \hat{k}$,$\vec{c} = 5 \hat{i}-\hat{j}+\lambda \hat{k}$,and $\vec{d} = 7 \hat{i}-4 \hat{j}+7 \hat{k}$.The vectors representing the edges are:$\vec{AB} = \vec{b} - \vec{a} = (-1-1)\hat{i} + (-3+6)\hat{j} + (7-10)\hat{k} = -2\hat{i} + 3\hat{j} - 3\hat{k}$$\vec{AC} = \vec{c} - \vec{a} = (5-1)\hat{i} + (-1+6)\hat{j} + (\lambda-10)\hat{k} = 4\hat{i} + 5\hat{j} + (\lambda-10)\hat{k}$$\vec{AD} = \vec{d} - \vec{a} = (7-1)\hat{i} + (-4+6)\hat{j} + (7-10)\hat{k} = 6\hat{i} + 2\hat{j} - 3\hat{k}$The volume of the tetrahedron is given by $V = \frac{1}{6} |[\vec{AB} \ \vec{AC} \ \vec{AD}]|$.Given $V = 11$,we have $11 = \frac{1}{6} |\det(\vec{AB}, \vec{AC}, \vec{AD})|$.$66 = |\det \begin{bmatrix} -2 & 3 & -3 \ 4 & 5 & \lambda-10 \ 6 & 2 & -3 \end{bmatrix}|$.Calculating the determinant:$-2(5(-3) - 2(\lambda-10)) - 3(4(-3) - 6(\lambda-10)) - 3(4(2) - 6(5))$$= -2(-15 - 2\lambda + 20) - 3(-12 - 6\lambda + 60) - 3(8 - 30)$$= -2(5 - 2\lambda) - 3(48 - 6\lambda) - 3(-22)$$= -10 + 4\lambda - 144 + 18\lambda + 66$$= 22\lambda - 88$.Since $66 = |22\lambda - 88|$,we have $22\lambda - 88 = 66$ or $22\lambda - 88 = -66$.Case $1$: $22\lambda = 154 \Rightarrow \lambda = 7$.Case $2$: $22\lambda = 22 \Rightarrow \lambda = 1$.Given the options,$\lambda = 7$ is the correct value.

If the volume of a tetrahedron,whose vertices are with position vectors $\hat{i}-6 \hat{j}+10 \hat{k}$,$-\hat{i}-3 \hat{j}+7 \hat{k}$,$5 \hat{i}-\hat{j}+\lambda \hat{k}$ and $7 \hat{i}-4 \hat{j}+7 \hat{k}$ is $11$ cubic units,then the value of $\lambda$ is:

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