Question

Given the vertices $A(2,3,1),B(4,1,-2),C(6,3,7)$ & $D(-5,-4,8)$ of a tetrahedron. The length of the altitude drawn from the vertex $D$ is

• A. $7$
• B. $9$
• C. $11$
• D. $Noneofthese$

Answer 1

The correct option is C $11$

$ax+by+cz+d=0$

$\therefore$ For $A:2a+3b+c+d=0⟶\left(1\right)$

For $B:4a+b-2c+d=0⟶\left(2\right)$

For $C:6a+3b+7c+d=0⟶\left(3\right)$

Now, from(1) and (3),

$2a+3b+c+d=6a+3b+7c+d$

$a=\frac{-3}{2}c⟶\left(4\right)$

From(1) and (2), $2a+3b+c+d=4a+b-2c+d$

$\Rightarrow -3c+3b+c=-6c+b-2c$ [Put value of ${}^{\prime }{a}^{\prime }$]

$\Rightarrow -6c-2b=0\Rightarrow b=-3c⟶\left(5\right)$

From(1), $2a+3b+c+d=0$

$-3c-9c+c+d=0\Rightarrow d=11c⟶\left(6\right)$

Now, $\frac{-3}{2}cx-3cy+cz+11c=0$

$\Rightarrow -3x-6y+2z+22=0[\because C\ne 0]$

$\Rightarrow 3x+6y-2z-22=0$

Length of $\perp$ar drawn from D to plane $ABC$,

$=\frac{|3(-5)+6(-4)+(-2)8-22|}{\sqrt{3^2+6^2+{(-2)}^2}}$

$=\frac{|-15-24-16-22|}{\sqrt{49}}=\frac{|-77|}{7}$

$=\frac{77}{7}=11$