Let the line L pass through the point (-3, 5, 2) and make equal angles with the positive coordinate

Q: Wait, why do we get two values for $r$? Does the line intersect the point twice?

Great observation! The set of all points at a fixed distance from a line forms a cylinder. The plane $y = r$ (or rather, the locus of points with a specific y-coordinate) intersects this cylinder at two distinct points, which is why we get two valid solutions for $r$. Both are physically and mathematically correct.

Q: Can I use the projection method instead of the cross product?

Absolutely! You can find the projection of $\vec{PQ}$ onto $\vec{d}$, find the foot of the perpendicular, and then use the distance formula between two points. However, the cross-product method is generally faster and less prone to arithmetic errors in competitive exams like JEE Advanced.

Q: Is the direction vector $\vec{d} = (1, 1, 1)$ unique?

Not at all! Any vector parallel to $(1, 1, 1)$, such as $(2, 2, 2)$ or $(-1, -1, -1)$, would work perfectly. The magnitude of the direction vector in the denominator of the distance formula scales proportionally, so the final result remains invariant. We choose $(1, 1, 1)$ simply for algebraic convenience.

Question

Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :

Options:

(A) 10

(B) 6

(C) 12

(D) 16

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system. You have a line $L$ that is perfectly symmetric, making equal angles with the $x, y,$ and $z$ axes.

Because the line makes equal angles $\alpha = \beta = \gamma$ with the axes, its direction cosines must satisfy the identity:

$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$

Since the angles are equal, we have $3 \cos^2 \alpha = 1$, which leads us to $\cos \alpha = \frac{1}{\sqrt{3}}$. This gives us the direction vector $\vec{d} = (1, 1, 1)$, which serves as the backbone of our line.

### Connecting the Dots

We have a fixed point $P(-3, 5, 2)$ on the line and an external point $Q(-2, r, 1)$. We need to find the distance between them.

The vector connecting these two points is calculated as:

$$\vec{PQ} = Q - P = (-2 - (-3))\hat{i} + (r - 5)\hat{j} + (1 - 2)\hat{k} = \hat{i} + (r - 5)\hat{j} - \hat{k}$$

This vector $\vec{PQ}$ represents the displacement from our line to the point $Q$.

### The Power of the Cross Product

To find the perpendicular distance, we use the vector formula:

$$d = \frac{|\vec{PQ} 	imes \vec{d}|}{|\vec{d}|}$$

The numerator, $|\vec{PQ} 	imes \vec{d}|$, represents the area of the parallelogram formed by $\vec{PQ}$ and $\vec{d}$. We calculate the cross product as follows:

$$\vec{PQ} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & r-5 & -1 \ 1 & 1 & 1 \end{vmatrix}$$

Expanding this determinant, we obtain:

$$\vec{PQ} 	imes \vec{d} = (r - 4)\hat{i} - 2\hat{j} + (6 - r)\hat{k}$$

### The Final Stretch

We are given the distance $d = \sqrt{\frac{14}{3}}$. Squaring both sides, we get:

$$d^2 = \frac{14}{3} = \frac{|\vec{PQ} 	imes \vec{d}|^2}{|\vec{d}|^2}$$

The magnitude squared of the cross product is $(r - 4)^2 + (-2)^2 + (6 - r)^2 = 2r^2 - 20r + 56$. The magnitude squared of $\vec{d}$ is $1^2 + 1^2 + 1^2 = 3$.

Equating these, we get:

$$\frac{14}{3} = \frac{2r^2 - 20r + 56}{3}$$

The denominators cancel out, leaving $14 = 2r^2 - 20r + 56$, which simplifies to the quadratic equation:

$$r^2 - 10r + 21 = 0$$

Factoring this quadratic gives $(r - 3)(r - 7) = 0$, so $r = 3$ or $r = 7$. The **sum of these values is $3 + 7 = 10$**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the line $L$ pass through the point $(- 3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(- 2, r, 1)$ is $\frac{14}{3}$ , then the sum of all possible values of $r$ is :

Select Answer:

Visualized Solution

Direction Vector of Line $L$

Points on the Line and Space

Position Vector $P Q$

Vector Distance Formula

Calculating $P Q \times d$

Expanding the Determinant

Magnitude Squared $∣ P Q \times d ∣^{2}$

Magnitude Squared $∣ d ∣^{2}$

Equating the Distances

Solving for $r$

Sum of Possible Values

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Connecting the Dots

The Power of the Cross Product

The Final Stretch

Similar Questions

Let a line pass through two distinct points $P (- 2, - 1, 3)$ and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ If the distance of the point Q from the point $R (1, 3, 3)$ is 5, then the square of the area of $△ P QR$ is equal to:

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$ . The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line $L$ along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to

If the length of the perpendicular drawn from the point $P (a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $26$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point $P$ in this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to :

Let the line passing through the points $(- 1, 2, 1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x+2}{3}= rac{y-3}{2}= rac{z-4}{1}$ at the point P. Then the distance of P from the point $Q (4, - 5, 1)$ is

Let a line $L$ pass through the point $P (2, 3, 1)$ and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z$ . If the distance of $L$ from the point $(5, 3, 8)$ is $α$ , then $3 α^{2}$ is equal to ______.

If the distances of the point $(1, 2, a)$ from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and $L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to

The distance of the point $P (4, 6, - 2)$ from the line passing through the point $(- 3, 2, 3)$ and parallel to a line with direction ratios $3, 3, - 1$ is equal to :

If a point R(4,y,z) lies on the line segment joining the points P(2,-3,4) and Q(8,0,10), then the distance of R from the origin is :

Let $P$ be the point $(10, - 2, - 1)$ and $Q$ be the foot of the perpendicular drawn from the point $R (1, 7, 6)$ on the line passing through the points $(2, - 5, 11)$ and $(- 6, 7, - 5)$ . Then the length of the line segment $P Q$ is equal to ________