Question

The ratio in which the XOZ plane divides the line segment joining the (1,-1,5) and (2,3,4) is
[a] -3:1
[b] -1:3
[c] 3:1
[d] 1:3

Answer 1

Hint: Assume that the ratio in which the xz plane divides the line segment is k:1. Hence find the coordinates of the point using the section formula. Use the fact that any point in the xz plane will have y-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 xz plane divides the given line segment.

Complete step-by-step solution
Let the ratio in which the xz plane divides the line segment joining the points A(1,-1,5) and B(2,3,4) 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, $C\equiv \left( \dfrac{2k+1}{k+1},\dfrac{3k-1}{k+1},\dfrac{4k+5}{k+1} \right)$
Since C lies on the xz plane, y-coordinate of C is 0
Hence, we have
$\dfrac{3k-1}{k+1}=0$
Multiplying both sides by k+1, we get
3k-1=0
Adding 1 on both sides, we get
3k = 1
Dividing both sides by 3, we get
$k=\dfrac{1}{3}$
Hence, the ratio in which the xz plane divides the line segment joining (1,-1,5) and (2,3,4) is 1:3
Hence we conclude that option [d] is correct

Note: Verification:
We can verify our solution by checking that the point which divides the line segment AB in the ratio 1:3 lies on the xz plane.
By section formula, we have
$C\equiv \left( \dfrac{2\times 1+1\times 3}{1+3},\dfrac{3\times 1-1\times 3}{1+3},\dfrac{4\times 1+5\times 3}{1+3} \right)=\left( \dfrac{5}{4},0,5 \right)$
Since the y-coordinate of C is 0, C lies on the xz plane.
Hence our solution is verified to be correct.