Question

The ratio in which the yz plane divides the line segment joining the (-3,4,2) and (2,1,3) is
[a] -4:1
[b] 3:2
[c] -2:3
[d] 1:4

Answer 1

Hint: Assume that the ratio in which the yz plane divides the line segment is k:1. Hence find the coordinates of the point using section formula. Use the fact that any point in the yz plane will have x-coordinate as 0. Hence form an equation in k. Solve for k. The value of k gives the ratio in which the point in the yz plane divides the given line segment.
Complete step by step solution:
Let the ratio in which the yz plane divides the line segment joining the points A(-3,4,2) and B(2,1,3) be k:1 at Point C

We know that if point C divides a line segment joining $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ in the ratio of m:n, then $C\equiv \left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$
Hence, $C\equiv \left( \dfrac{2k-3}{k+1},\dfrac{k+4}{k+1},\dfrac{3k+2}{k+1} \right)$
Since C lies on the yz plane, x-coordinate of C is 0
Hence, we have
$\dfrac{2k-3}{k+1}=0$
Multiplying both sides by k+1, we get
2k-3=0
Adding 3 on both sides, we get
2k = 3
Dividing both sides by 2, we get
$k=\dfrac{3}{2}$
Hence, the ratio in which the yz plane divides the line segment joining (-3,4,2) and (2,1,3) is 3:2
Hence option [b] is correct
Note: Verification:
We can verify our solution by checking that the point which divides the line segment AB in the ratio 3:2 lies on the yz plane.
By section formula, we have
$C\equiv \left( \dfrac{3\times 2-3\times 2}{3+2},\dfrac{3\times 1+4\times 2}{3+2},\dfrac{3\times 3+2\times 2}{3+2} \right)=\left( 0,\dfrac{11}{5},\dfrac{13}{5} \right)$
Since x-coordinate of C is 0, C lies on the yz plane.
Hence our solution is verified to be correct.