Let the co-ordinates of one vertex of ∆ ABC be A ( 0, 2, α ) and the other two vertices lie on the line (x + α)/(5) = (y - 1)/(2) = (z + 4)/(3) . For α ∈ ℤ , if the area of ∆ ABC is 21sq . units and the line segment BC

Correct answer: 0 Given, The co-ordinates of one vertex of ∆ ABC be A ( 0, 2, α ) and the other two vertices lie on the line (x + α)/(5) = (y - 1)/(2) = (z + 4)/(3) Now plotting the diagram we get, Now any point on the plane (x + α)/(5) = (y - 1)/(2) = (z + 4)/(3) = k is given by (5k - α, 2k + 1, 3k - 4) Now direction ratio of line AD is given by, (5k - α - 0, 2k + 1 - 2, 3k - 4 - α) ≡ (5k - α, 2k - 1, 3k - 4 - α) Now AD is perpendicular to BC , So, 5(5k - α) + 2(2k - 1) + 3(3k - 4 - α) = 0 ⇒ 19k - 4α - 7 = 0 . . . . . (1) Also given area, 21 = (1)/(2) × 2 21 × AD ⇒ AD = 21 ⇒ (5k - α) + (2k - 1) + (3k - 4 - α) = 21 ⇒ 19 k ^ 2 - 8kα + α ^ 2 - 14k + 4α = 2 . . . . . . (2) Now on solving equation (1) and (2) we get, α = 3 ⇒ α ^ 2 = 9 .

JEE Main 202329 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2023 · 29 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let the co-ordinates of one vertex of ∆ ABC be A ( 0, 2, α ) and...

Let the co-ordinates of one vertex of $∆ ABC$ be $A ( 0, 2, \alpha )$ and the other two vertices lie on the line $\dfrac{x + \alpha}{5} = \dfrac{y - 1}{2} = \dfrac{z + 4}{3}$ . For $\alpha \in \mathbb{Z}$ , if the area of $∆ ABC$ is $21sq . units$ and the line segment $BC$ has length $2\sqrt{21}$ units, then ${\alpha}^{2}$ is equal to _______.

Correct answer

Step-by-step solution

Given,

The co-ordinates of one vertex of $∆ ABC$ be $A ( 0, 2, \alpha ) and the other two vertices lie on the line$ \dfrac{x + \alpha}{5} = \dfrac{y - 1}{2} = \dfrac{z + 4}{3} $Now plotting the diagram we get,<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/5852485c-e812-4825-ba42-7bbd4a508d66-9894843_re.PNG" style="width: 346px; height: 150px;" />Now any point on the plane$ \dfrac{x + \alpha}{5} = \dfrac{y - 1}{2} = \dfrac{z + 4}{3} = k $is given by \left(5k - \alpha, 2k + 1, 3k - 4\right)$

Now direction ratio of line $AD$ is given by, \left(5k - \alpha - 0, 2k + 1 - 2, 3k - 4 - \alpha\right) $$ ≡ \left(5k - \alpha, 2k - 1, 3k - 4 - \alpha\right) $Now$ AD $is perpendicular to$ BC $,So,$ 5\left(5k - \alpha\right) + 2\left(2k - 1\right) + 3\left(3k - 4 - \alpha\right) = 0 $$ ⇒ 19k - 4\alpha - 7 = 0 . . . . . \left(1\right) $Also given area,$ 21 = \dfrac{1}{2} \times 2\sqrt{21} \times AD $$ ⇒ AD = \sqrt{21} $$ ⇒ \left(5k - \alpha\right) + \left(2k - 1\right) + \left(3k - 4 - \alpha\right) = 21 $$ ⇒ 19{k}^{2} - 8k\alpha + {\alpha}^{2} - 14k + 4\alpha = 2 . . . . . . \left(2\right) $Now on solving equation \left(1\right) \text{ and } \left(2\right)$ we get,

$\alpha = 3 ⇒ {\alpha}^{2} = 9$ .

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