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JEE Main 202315 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
15 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2023 · 15 April · Morning Shift · Mathematics · Three Dimensional Geometry

Let the foot of perpendicular of the point P(3, – 2, – 9) on the...

Let the foot of perpendicular of the point P(3,–2,–9)onP\left(3, – 2, – 9\right) on the planepassingthroughthepoints(–1,–2,–3),(9,3,4),(9,–2,1)beplane passing through the points \left(– 1, – 2, – 3\right), \left(9, 3, 4\right), \left(9, – 2, 1\right) beQ\left(\alpha, \beta, \gamma\right).Thenthedistance. Then the distanceQ$ from the origin isLet the foot of perpendicular of the point P(3, – 2, – 9) on the plane passing through the points (– 1, – 2, – 3), (9,

Options

  1. A42\sqrt{42}
  2. B38\sqrt{38}
  3. C35\sqrt{35}
  4. D29\sqrt{29}

Correct answer

A. 42\sqrt{42}

Step-by-step solution

Given that three points lie on a plane,

Let the foot of perpendicular of the point P(3, – 2, – 9) on the plane passing through the points (– 1, – 2, – 3), (9,

\overset{\to}{AC} = 10\overset{^}{i} + 4\overset{^}{k}

\overset{\to}{AB} = 10\overset{^}{i} + 5\overset{^}{j} + 7\overset{^}{k}

\overset{\to}{AC} \times \overset{\to}{AB} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 10 & 0 & 4 \\ 10 & 5 & 7\end{cases}\right|

⇒ \overset{\to}{AC} \times \overset{\to}{AB} = - 20\overset{^}{i} - 30\overset{^}{j} + 50\overset{^}{k}

⇒ \overset{\to}{AC} \times \overset{\to}{AB} = 2\overset{^}{i} + 3\overset{^}{j} - 5\overset{^}{k}

Equation of plane will be,

⇒2x+3y−5z=d⇒ 2x + 3y - 5z = d

And the plane passing through, \left(– 1, – 2, – 3\right)</p><p></p><p>⇒ - 2 - 6 + 15 = d</p><p></p><p>⇒ d = 7</p><p></p><p>∴ 2x + 3y - 5z = 7</p><p>Nowfindingthefootofperpendicularweget,</p><p></p><p>Now finding the foot of perpendicular we get,</p><p>\dfrac{x - 3}{2} = \dfrac{y + 2}{3} = \dfrac{z + 9}{- 5} = - \left(\dfrac{38}{38}\right)</p><p></p><p>x = 1, y = - 5, z = - 4</p><p></p><p>Q\left(1, – 5, – 4\right)</p><p>Distancefromorigin</p><p>Distance from origin= \sqrt{1 + 25 + 16}$$= \sqrt{42}

Hence this is the required option.

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