Question:

Let a unit vector $ \mathbf{OP} $ make angles $ \alpha, \beta, \gamma $ with the positive directions of the coordinate axes $ OX, OY, OZ $, respectively, where $ \beta \in [0, \frac{\pi}{2}] $, and $ \mathbf{OP} $ is perpendicular to the plane through the points $ (1,2,3) $, $ (2,3,4) $, and $ (1,5,7) $. Then which one of the following is true?

Updated On: Jan 8, 2025

$\alpha \in\left(0, \frac{\pi}{2}\right)$ and $\gamma \in\left(0, \frac{\pi}{2}\right)$
$\alpha \in\left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in\left(\frac{\pi}{2}, \pi\right)$
$\alpha \in\left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in\left(0, \frac{\pi}{2}\right)$
$\alpha \in\left(0, \frac{\pi}{2}\right)$ annd $\gamma \in\left(\frac{\pi}{2}, \pi\right)$

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The Correct Option is B

Solution and Explanation

We are given three points $(1, 2, 3)$, $(2, 3, 4)$, and $(1, 5, 7)$, and we need to find the angle that the unit vector $\overrightarrow{OP}$ makes with the coordinate axes.

Step 1: Equation of the Plane

The equation of the plane can be determined using the determinant of a matrix formed from the coordinates of the points:

\[ \begin{vmatrix} x - 1 & y - 2 & z - 3 \\ 1 & 1 & 0 \\ 3 & 1 & 4 \end{vmatrix} = 0. \]

Expanding this determinant gives:

\[ (x - 1) \cdot 4 - (y - 2) \cdot 3 + (z - 3) \cdot 2 = 0. \]

Simplify:

\[ x - 4y + 3z = 2. \]

Thus, the equation of the plane is:

\[ x - 4y + 3z = 2. \]

Step 2: Direction Ratios of the Normal to the Plane

The direction ratios of the normal to the plane are $\langle 1, -4, 3 \rangle$.

Step 3: Direction Cosines of the Normal Vector

The direction cosines of the normal vector are given by:

\[ \cos\alpha = \frac{-1}{\sqrt{26}}, \quad \cos\beta = \frac{4}{\sqrt{26}}, \quad \cos\gamma = \frac{-3}{\sqrt{26}}. \]

Step 4: Finding the Angles

The angles $\alpha$, $\beta$, and $\gamma$ correspond to the direction cosines:

$\cos\beta = \frac{4}{\sqrt{26}}$, so $\beta \in [0, \frac{\pi}{2}]$.
$\cos\alpha = \frac{-1}{\sqrt{26}}$, so $\alpha \in [\frac{\pi}{2}, \pi]$.
$\cos\gamma = \frac{-3}{\sqrt{26}}$, so $\gamma \in [\frac{\pi}{2}, \pi]$.

Conclusion

The correct answer is:

\[ \boxed{1}. \]

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Top Questions on Three Dimensional Geometry

Let y ∈ R be such that the lines $L_1:\frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2:\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ intersect. Let R₁ be the point of intersection of L₁ and L₂. Let O = (0, 0 ,0), and $\hat{n}$ denote a unit normal vector to the plane containing both the lines L₁ and L₂.
Match each entry in List-I to the correct entry in List-II.

List - I		List - II
(P)	γ equals	(1)	$-\hat{i}-\hat{j}+\hat{k}$
(Q)	A possible choice for $\hat{n}$ is	(2)	$\sqrt{\frac{3}{2}}$
(R)	$\overrightarrow{OR_1}$ equals	(3)	1
(S)	A possible value of $\overrightarrow{OR_1}.\hat{n}$ is	(4)	$\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}$
		(5)	$\sqrt{\frac{2}{3}}$

The correct option is

JEE Advanced - 2024
Mathematics
Three Dimensional Geometry

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Concepts Used:

Three Dimensional Geometry

Mathematically, Geometry is one of the most important topics. The concepts of Geometry are derived w.r.t. the planes. So, Geometry is divided into three major categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.

Direction Cosines and Direction Ratios of Line:

Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.