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Find x such that the four points A(5, x, 4), B(4, 4, 0), C(5, 4, – 3) and D(7, 7, – 2) are coplanar.

Answers (1)

$A=(5,x,4), B=(4,4,0), C=(5,4,-3), D=(7,7,2)$

Four points A,B,C,D are coplanar if the three vectors $\overrightarrow{AB},\overrightarrow{AC}\: and\: \overrightarrow{AD}$ are coplanar.

i.e $\left [ \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD} \right ]=0$

$\overrightarrow{AB}=(4-5)\hat{i}+(4-x)\hat{j}+(-4)\hat{k}$

$=-1\hat{i}+(4-x)\hat{j}-4\hat{k}$

$\overrightarrow{AC}=(5-5)\hat{i}+(4-x)\hat{j}+(-3-4)\hat{k}$

$=0\hat{i}+(4-x)\hat{j}+(-7)\hat{k}$

$\overrightarrow{AD}=(7-5)\hat{i}+(7-x)\hat{j}+(-2-4)\hat{k}$

$=2\hat{i}+(7-x)\hat{j}-6\hat{k}$

$\therefore \begin{bmatrix} \overrightarrow{AB} &\overrightarrow{AC} &\overrightarrow{AD} \end{bmatrix}=\begin{vmatrix} -1 &4-x &-4 \\ 0 &4-x &-7 \\ 2 &7-x & -6 \end{vmatrix}\\\Rightarrow 0=-1[(4-x)(-6)-(7-x)(-7)]-(4-x)[0+14]+(-4)[0-2(4-x)]$

$\Rightarrow 0=-1[-24+6x+49-7x]-56+14x-4[-8+2x]$

$\Rightarrow 0=-1[25-x]-56+14x+32-8x$

$\Rightarrow 0=-25+x-56+14x+32-8x$

$\Rightarrow 0=7x-49$

$\Rightarrow 7x=49$

$\Rightarrow x=\frac{49}{7}=7$

Posted by

Ravindra Pindel

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