Three Dimensional Geometry jee mains questions with solutions

(3)

$M is the point of intersection of L_{1} and L_{2}$

$\Rightarrow 2 λ - 1 = 2 μ - 1, 3 λ - 1 = 3 μ - 1, 6 λ - 3 = 9$

$\Rightarrow λ = 2 = μ$

$\Rightarrow M (3, 5, 9)$

$Now let point P be (2 K - 1, 3 K - 1, 6 K - 3) on L_{2},$ $such that P M = 7$

$\Rightarrow \sqrt{{(2 K - 4)}^{2} + {(3 K - 6)}^{2} + {(6 K - 12)}^{2}} = 7$

$\Rightarrow 49 K^{2} + 196 - 196 K = 49$

$\Rightarrow K^{2} + 4 - 4 K = 1$

$\Rightarrow K^{2} - 4 K + 3 = 0$

$\Rightarrow K = 1, 3$

$So points P and Q are (1, 2, 3) and (5, 8, 15)$

$So sum of all coordinates of P and Q = 34$

(3)

$L : \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$

$L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b} = λ$

$L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c} = μ$

$Let A (3 λ + 1, 4 λ + 2, b λ + a)$

$It lies on L$

$∴$ $\frac{3 λ}{1} = \frac{4 λ + 2}{2} = \frac{b λ + a - 1}{1}$

$\Rightarrow λ = 1 and a + b - 1 = 3$

$\Rightarrow A (4, 6, 4), a + b = 4 . . . (1)$

$Let B (μ + 1, 4 μ + 2, c μ + a)$

$It also lies on L$

$\frac{μ}{1} = \frac{4 μ + 2}{2} = \frac{c μ + a - 1}{1}$

$\Rightarrow 2 μ = 4 μ + 2$

$\Rightarrow μ = - 1$

$a - c - 1 = - 1$

$\Rightarrow a = c . . . (2), B (0, - 2, 0)$

$Also P A = P B, P (1, 2, a), A (4, 6, 4)$

$\Rightarrow 9 + 16 + {(a - 4)}^{2} = 1 + 16 + a^{2}$

$\Rightarrow 16 + 8 = 8 a$

$\Rightarrow a = 3$ $∴$ $c = 3, b = 1$

$∴$ $a + b + c = 7$

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