**Question:**

Find the ratio in which the plane x – 2y + 3z = 5 divides the join of A(3, -5, 4) and B(2, 3, -7). Find the coordinates of the point of intersection of the line and the plane.

**Solution:**

Let the plane x – 2y + 3z = 5 divides the join of A(3, -5, 4) and B(2, 3, -7) in ratio k:1.

The point which will come by section formula will be in the plane. Putting that in the plane equation will give the point coordinates. The points are $A(3,-5,4)$ and $B(2,3,-7)$.

Using section formula,

$\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{mz}_{2}+\mathrm{nz}_{1}}{\mathrm{~m}+\mathrm{n}}\right)$

we get

$=\left(\frac{\mathrm{k} \times 2+1 \times 3}{\mathrm{k}+1}, \frac{\mathrm{k} \times 3+1 \times-5}{\mathrm{k}+1}, \frac{\mathrm{k} \times-7+1 \times 4}{\mathrm{k}+1}\right)$

Putting this point in the plane equation, we get

$\frac{2 k+3}{k+1}-2\left(\frac{3 k-5}{k+1}\right)+3\left(\frac{-7 k+4}{k+1}\right)=5$

$2 k+3-6 k+10-21 k+12=5 k+5$

$-25 k+25=5 k+5$

$-30 k=-20$

$\mathrm{k}=\frac{2}{3}$

the ratio is $2: 3$. And the point of intersection of the plane and the line is

$\left(\frac{13}{5},-\frac{9}{5},-\frac{2}{5}\right)$

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