Consider the four points A(1, -2, -1), B(4, 0 | vedclass

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(D) Given points are $A(1, -2, -1)$, $B(4, 0, -3)$, $C(1, 2, -1)$, and $D(2, -4, -5)$.The vectors are:$\vec{b} = \vec{AB} = (4-1)\hat{i} + (0-(-2))\ha

Vedclass · Accepted Answer

$-20$

Vedclass · Answer

(D) Given points are $A(1, -2, -1)$, $B(4, 0, -3)$, $C(1, 2, -1)$, and $D(2, -4, -5)$.The vectors are:$\vec{b} = \vec{AB} = (4-1)\hat{i} + (0-(-2))\hat{j} + (-3-(-1))\hat{k} = 3\hat{i} + 2\hat{j} - 2\hat{k}$$\vec{c} = \vec{AC} = (1-1)\hat{i} + (2-(-2))\hat{j} + (-1-(-1))\hat{k} = 0\hat{i} + 4\hat{j} + 0\hat{k}$$\vec{d} = \vec{AD} = (2-1)\hat{i} + (-4-(-2))\hat{j} + (-5-(-1))\hat{k} = 1\hat{i} - 2\hat{j} - 4\hat{k}$The scalar triple product $[\vec{b}, \vec{c}, \vec{d}] = \begin{vmatrix} 3 & 2 & -2 \ 0 & 4 & 0 \ 1 & -2 & -4 \end{vmatrix} = 3(-16) - 2(0) - 2(-4) = -48 + 8 = -40$.We know that $[\vec{b} 	imes \vec{c}, \vec{c} 	imes \vec{d}, \vec{d} 	imes \vec{b}] = [\vec{b}, \vec{c}, \vec{d}]^2 = (-40)^2 = 1600$.Also, $[\vec{b}+\vec{c}, \vec{c}+\vec{d}, \vec{d}+\vec{b}] = 2[\vec{b}, \vec{c}, \vec{d}] = 2(-40) = -80$.Therefore, the required value is $\frac{1600}{-80} = -20$.

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