Let the coordinates of one vertex of triangle | vedclass

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(C) The area of $	riangle ABC$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 21$. Given the base $BC = 2\sqrt{21}$,the height $h

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) The area of $	riangle ABC$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 21$. Given the base $BC = 2\sqrt{21}$,the height $h$ (perpendicular distance from $A$ to the line) is $\frac{2 	imes 21}{2\sqrt{21}} = \sqrt{21}$. The line is $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3} = k$. The direction vector is $\vec{v} = 5\hat{i} + 2\hat{j} + 3\hat{k}$. $A$ point on the line is $P(-\alpha, 1, -4)$. The vector $\vec{AP} = (-\alpha - 0)\hat{i} + (1 - 2)\hat{j} + (-4 - \alpha)\hat{k} = -\alpha\hat{i} - \hat{j} - (\alpha + 4)\hat{k}$. The perpendicular distance $h = \frac{|\vec{AP} 	imes \vec{v}|}{|\vec{v}|}$. $\vec{AP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -\alpha & -1 & -(\alpha+4) \ 5 & 2 & 3 \end{vmatrix} = \hat{i}(-3 + 2\alpha + 8) - \hat{j}(-3\alpha + 5\alpha + 20) + \hat{k}(-2\alpha + 5) = (2\alpha + 5)\hat{i} - (2\alpha + 20)\hat{j} + (5 - 2\alpha)\hat{k}$. $|\vec{AP} 	imes \vec{v}|^2 = (2\alpha + 5)^2 + (2\alpha + 20)^2 + (5 - 2\alpha)^2 = 4\alpha^2 + 20\alpha + 25 + 4\alpha^2 + 80\alpha + 400 + 25 - 20\alpha + 4\alpha^2 = 12\alpha^2 + 80\alpha + 450$. Since $h^2 = 21$ and $|\vec{v}|^2 = 5^2 + 2^2 + 3^2 = 38$,we have $\frac{12\alpha^2 + 80\alpha + 450}{38} = 21$. $12\alpha^2 + 80\alpha + 450 = 798 \Rightarrow 12\alpha^2 + 80\alpha - 348 = 0 \Rightarrow 3\alpha^2 + 20\alpha - 87 = 0$. Solving for $\alpha$: $\alpha = \frac{-20 \pm \sqrt{400 - 4(3)(-87)}}{6} = \frac{-20 \pm \sqrt{400 + 1044}}{6} = \frac{-20 \pm 38}{6}$. Taking $\alpha = 3$,we get $\alpha^2 = 9$.

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