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Question: A point on XOZ-plane divides the join of \( (5, - 3, - 2) \) and \( (1,2, - 2) \) at A. \( (\dfrac...

A point on XOZ-plane divides the join of (5,−3,−2)(5, - 3, - 2) and (1,2,−2)(1,2, - 2) at
A. (135,0,−2)(\dfrac{{13}}{5},0, - 2)
B. (135,0,2)(\dfrac{{13}}{5},0,2)
C. (5,0,2)(5,0,2)
D. (5,0,−2)(5,0, - 2)

Explanation

Solution

Hint : A plane is a surface formed by a straight line joining two points such that it will lie wholly on it that is the surface formed by this line when it moves parallel to itself in both the horizontal or vertical directions. The plane formed by the combination of x-axis and y-axis is called X0Z-plane; any point lying on this plane will have its y-coordinate zero. Using this information, we can find out the correct answer.

Complete step-by-step answer :
The equation of the line joining the points (5,−3,−2)(5, - 3, - 2) and (1,2,−2)(1,2, - 2) is given as –
r⃗=a⃗+λ(b⃗−a⃗) r⃗=i^+2j^−2k^+λ[5i^−3j^−2k^−(i^+2j^−2k^)] r⃗=i^+2j^−2k^+λ[(5−1)i^+(−3−2)j^+(−2+2)k^] r⃗=i^+2j^−2k^+λ(4i^−5j^) r⃗=i^+4λi^+2j^−5λj^−2k^ r⃗=(1+4λ)i^+(2−5λ)j^−2k^   \vec r = \vec a + \lambda (\vec b - \vec a) \\\ \vec r = \hat i + 2\hat j - 2\hat k + \lambda [5\hat i - 3\hat j - 2\hat k - (\hat i + 2\hat j - 2\hat k)] \\\ \vec r = \hat i + 2\hat j - 2\hat k + \lambda [(5 - 1)\hat i + ( - 3 - 2)\hat j + ( - 2 + 2)\hat k] \\\ \vec r = \hat i + 2\hat j - 2\hat k + \lambda (4\hat i - 5\hat j) \\\ \vec r = \hat i + 4\lambda \hat i + 2\hat j - 5\lambda \hat j - 2\hat k \\\ \vec r = (1 + 4\lambda )\hat i + (2 - 5\lambda )\hat j - 2\hat k \;
Now, we have to find the point on the XOZ-plane which divides the given line, that point will satisfy the equation of both the plane and the line. Y-coordinate at any point on the XOZ-plane is zero, so the y-coordinate at that point on the line will also be zero. That is-
2−5λ=0 ⇒λ=25   2 - 5\lambda = 0 \\\ \Rightarrow \lambda = \dfrac{2}{5} \;
The position vector of that point will be
⇒r⃗=[1+4(25)]i^+[2−5(25)]j^−2k^ ⇒r⃗=135i^+0j^−2k^   \Rightarrow \vec r = [1 + 4(\dfrac{2}{5})]\hat i + [2 - 5(\dfrac{2}{5})]\hat j - 2\hat k \\\ \Rightarrow \vec r = \dfrac{{13}}{5}\hat i + 0\hat j - 2\hat k \;
Thus the coordinates of the point on the XOZ-plane that divides the join of points (5,−3,−2)(5, - 3, - 2) and (1,2,−2)(1,2, - 2) is (135,0,−2)(\dfrac{{13}}{5},0, - 2) .
So, the correct answer is “Option A”.

Note : The equation of a line through two passing points is given as –
r⃗=a⃗+λ(b⃗−a⃗)\vec r = \vec a + \lambda (\vec b - \vec a)
So, we have to first convert the coordinates given in the question to the vector form, we know that a point having coordinates (a,b,c)(a,b,c) is written in vector form as ai^+bj^+ck^a\hat i + b\hat j + c\hat k .
So, a⃗=i^+2j^−2k^\vec a = \hat i + 2\hat j - 2\hat k and b⃗=5i^−3j^−2k^\vec b = 5\hat i - 3\hat j - 2\hat k in the given question.