A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is

Q: Wait, why can't we just find the coordinates of the projection point $P'$ and then find the distance $OP'$?

You absolutely can! That is a perfectly valid geometric approach. However, finding the coordinates of $P'$ requires finding the intersection of the line passing through $P$ perpendicular to the plane, which involves solving for a parameter $t$ and substituting it back into the line equation. It is a longer, more calculation-heavy process. The vector decomposition method we used is faster and significantly less prone to arithmetic errors.

Q: Wait, why do we use the normal vector $\vec{n}$ to find the projection? Isn't the projection on the plane?

That is a brilliant question. The normal vector $\vec{n}$ is perpendicular to the plane. By projecting $\vec{OP}$ onto $\vec{n}$, we are essentially finding the 'height' of the vector relative to the plane. Since $\vec{OP}$ is the hypotenuse of a right triangle where one leg is this 'height' (the normal component) and the other leg is the projection on the plane, we can use the Pythagorean theorem to find the length of the projection on the plane without needing to find the coordinates of the projected point.

Q: Is the projection of a vector always a vector? Why are we finding the length?

Yes, the projection of a vector onto a plane is indeed a vector. However, the question specifically asks for the *length* of this projection. Finding the length directly using the Pythagorean relationship is much more efficient than finding the vector components first. In JEE Advanced, recognizing when you can solve for a scalar magnitude instead of the full vector is a key skill for saving time.

Question

A plane passes through the points $A(1, 2, 3), B(2, 3, 1)$ and $C(2, 4, 2)$. If $O$ is the origin and $P$ is $(2, -1, 1)$, then the projection of $\vec{OP}$ on this plane is of length:

Options:

(A) $\sqrt{\frac{2}{5}}$

(B) $\sqrt{\frac{2}{3}}$

(C) $\sqrt{\frac{2}{11}}$

(D) $\sqrt{\frac{2}{7}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have three points, $A(1, 2, 3)$, $B(2, 3, 1)$, and $C(2, 4, 2)$, floating in space. These three points define a unique, flat surface—a plane.

Now, you have the origin $O(0, 0, 0)$ and a point $P(2, -1, 1)$ somewhere else. You are asked to find the length of the projection of the vector $\vec{OP}$ onto this plane.

### The Soul of the Plane

Before we can project anything, we need to define the plane itself. The most elegant way to find the equation of the plane is the determinant method.

We consider a general point $(x, y, z)$ on the plane and form vectors with the given points. The condition that these vectors are coplanar is that their scalar triple product is zero:

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

Simplifying the rows, we get:

$$
\begin{vmatrix} x - 1 & y - 2 & z - 3 \ 1 & 1 & -2 \ 1 & 2 & -1 \end{vmatrix} = 0
$$

Expanding this determinant along the first row, we obtain:

$$
(x - 1)[(1)(-1) - (2)(-2)] - (y - 2)[(1)(-1) - (1)(-2)] + (z - 3)[(1)(2) - (1)(1)] = 0
$$

This simplifies to $3(x - 1) - 1(y - 2) + 1(z - 3) = 0$. This gives us the plane equation: **$3x - y + z - 4 = 0$**.

The coefficients of $x, y,$ and $z$ give us the normal vector $\vec{n} = 3\hat{i} - \hat{j} + \hat{k}$. This normal vector is the compass of our plane; it tells us exactly how the plane is oriented in space.

### The Vector Decomposition Strategy

Now, we have our vector $\vec{OP} = 2\hat{i} - \hat{j} + \hat{k}$. We want to project this onto the plane.

Instead of finding the coordinates of the projection point, we use a powerful geometric insight. We can decompose $\vec{OP}$ into two orthogonal components: one parallel to the plane (our target projection) and one perpendicular to the plane (along the normal $\vec{n}$).

This forms a right-angled triangle where $\vec{OP}$ is the hypotenuse. The length of the component along the normal, $d$, is the projection of $\vec{OP}$ onto $\vec{n}$, calculated as:

$$
d = \frac{|\vec{OP} \cdot \vec{n}|}{|\vec{n}|}
$$

First, let us find the magnitudes: $|\vec{OP}|^2 = 2^2 + (-1)^2 + 1^2 = 6$ and $|\vec{n}|^2 = 3^2 + (-1)^2 + 1^2 = 11$, so $|\vec{n}| = \sqrt{11}$.

The dot product is $\vec{OP} \cdot \vec{n} = (2)(3) + (-1)(-1) + (1)(1) = 6 + 1 + 1 = 8$. Thus, $d = \frac{8}{\sqrt{11}}$.

### Final Calculation

Now, we use the Pythagorean theorem: $|\vec{OP}|^2 = L^2 + d^2$, where $L$ is the length of the projection on the plane. Rearranging, we get $L = \sqrt{|\vec{OP}|^2 - d^2}$.

Substituting our values:

$$
L = \sqrt{6 - \left(\frac{8}{\sqrt{11}}ight)^2} = \sqrt{6 - \frac{64}{11}} = \sqrt{\frac{66 - 64}{11}} = \sqrt{\frac{2}{11}}
$$

And there it is! The length of the projection is **$\sqrt{\frac{2}{11}}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: A plane passes through the points $A (1, 2, 3), B (2, 3, 1)$ and $C (2, 4, 2)$ . If $O$ is the origin and $P$ is $(2, - 1, 1)$ , then the projection of $O P$ on this plane is of length:

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Equation of a Plane through 3 Points

Substituting the Coordinates

Simplifying the Determinant

Expanding the Determinant

Final Plane Equation & Normal Vector

Defining Vector OP

The Projection Strategy

Calculating Magnitudes

Finding the Normal Component ( $d$ )

Calculating the Plane Projection ( $L$ )

Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Soul of the Plane

The Vector Decomposition Strategy

Final Calculation

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