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The sum of the intercepts on the coordinate axes of the plane passing through the point (−2, −2, 2) and containing the line joining the points (1, −1, 2) and (1, 1, 1), is :

Answers (1)

As we have learned

Cartesian equation of plane passing through a given point and normal to a given vector -

$\left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0$

- wherein

$\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}$

$\vec{a}= x_{0}\hat{i}+y_{0}\hat{j}+z_{0}\hat{k}$

$\vec{n}= a\hat{i}+b\hat{j}+c\hat{k}$

Putting in

$\left ( \vec{r}-\vec{a} \right )\cdot \vec{n}= 0$

We get $\left ( x-x_{0} \right )a+\left ( y-y_{0} \right )b+\left ( z-z_{0} \right )c= 0$

$a(x-x_{1})+b(y-y_{1})+c(z-z_{1})=0$

$a(x+2)+b(y+2)+c(z-2)=0$

passing through (1,-1,2)

and (1,1,1)

$\Rightarrow 3b+b+0c=0$

$\Rightarrow 3b+3b-c=0$

$\frac{a}{-1}=\frac{b}{3}=\frac{c}{6}$

$-x-2+3y+6+62-12=0$

$-x+3y+6z=8$

$\frac{x}{-8}+\frac{y}{8/3}+\frac{z}{4/3}=1$

$\large \begin{array}{l} =-8+\frac{8}{3}+\frac{8}{6} \\\\ =-8+\frac{16+8}{6} \\\\ =-8+\frac{24}{6} \\\\ =-8+4 \\\\ =-4 \end{array}$

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