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Mathematics Question on 3D Geometry

The distance of the point Q(0,2,−2)Q(0, 2, -2) from the line passing through the point P(5,−4,3)P(5, -4, 3) and perpendicular to the lines r⃗=(−3i^+2k^)+λ(2i^+3j^+5k^),λ∈R\vec{r} = (-3\hat{i} + 2\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 5\hat{k}), \quad \lambda \in \mathbb{R} and r⃗=(i^−2j^+k^)+μ(−i^+3j^+2k^),μ∈R\vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu (-\hat{i} + 3\hat{j} + 2\hat{k}), \quad \mu \in \mathbb{R} is

A

86\sqrt{86}

B

20\sqrt{20}

C

54\sqrt{54}

D

74\sqrt{74}

Answer

74\sqrt{74}

Explanation

Solution

A vector in the direction of the required line can be obtained by the cross product of:

∣i^j^k^ 235 −132∣=−9i^−9j^+9k^\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 2 & 3 & 5 \\\ -1 & 3 & 2 \end{vmatrix} = -9\hat{i} - 9\hat{j} + 9\hat{k}

Required line:

r⃗=(5i^−4j^+3k^)+λ(−9i^−9j^+9k^)\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda (-9\hat{i} - 9\hat{j} + 9\hat{k}) r⃗=(5i^−4j^+3k^)+λ(1i^+j^−k^)\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda (1\hat{i} + \hat{j} - \hat{k})

Now, the distance of (0,2,−2)(0, 2, -2) is:

P.V. of P=(5+λ)i^+(−4+λ)j^+(3−λ)k^\text{P.V. of } P = (5 + \lambda)\hat{i} + (-4 + \lambda)\hat{j} + (3 - \lambda)\hat{k} AP⃗=(5+λ)i^+(−6+λ)j^+(5−λ)k^\vec{AP} = (5 + \lambda)\hat{i} + (-6 + \lambda)\hat{j} + (5 - \lambda)\hat{k} AP⃗⋅(i^+j^−k^)=0\vec{AP} \cdot (\hat{i} + \hat{j} - \hat{k}) = 0 5+λ−6+λ−5+λ=0  ⟹  λ=25 + \lambda - 6 + \lambda - 5 + \lambda = 0 \quad \implies \quad \lambda = 2

∣AP⃗∣=(5+λ)2+(−6+λ)2+(5−λ)2|\vec{AP}| = \sqrt{(5 + \lambda)^2 + (-6 + \lambda)^2 + (5 - \lambda)^2} ∣AP⃗∣=49+16+9=74|\vec{AP}| = \sqrt{49 + 16 + 9} = \sqrt{74}