The distance of the point Q(0, 2, -2) form the line passing through the point P(5, -4, 3) and perpen

Q: Why do we take the cross product of the direction vectors? Isn't there a simpler way?

The cross product of two vectors $\vec{b}_1$ and $\vec{b}_2$ is the only operation that yields a vector mutually perpendicular to both. Since our required line must be perpendicular to both given lines, its direction vector must be parallel to this cross product. It is not just a method; it is the geometric definition of the orientation we need.

Q: Can I simplify the direction vector $\vec{b} = -9\hat{i} - 9\hat{j} + 9\hat{k}$ to $\hat{i} + \hat{j} - \hat{k}$? Won't that change the line?

Absolutely, you can and should! In vector geometry, the direction of a line is defined by its direction ratios. Any non-zero scalar multiple of a direction vector represents the same line. Simplifying by dividing by $-9$ makes your calculations significantly cleaner and drastically reduces the risk of arithmetic errors. It is a pro-move that separates the masters from the novices.

Q: Why does the dot product $\vec{QM} \cdot \vec{b} = 0$ work? I always forget why we set it to zero.

This is the fundamental condition for perpendicularity. If two non-zero vectors are perpendicular, the angle between them is $90^\circ$. Since the dot product is defined as $|\vec{A}||\vec{B}|\cos(\theta)$, and $\cos(90^\circ) = 0$, the entire product must vanish. This is the most powerful tool in your arsenal for finding the foot of a perpendicular.

Question

The distance of the point $Q(0, 2, -2)$ form the line passing through the point $P(5, -4, 3)$ and perpendicular to the lines $\vec{r} = (-3\hat{i} + 2\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 5\hat{k}), \lambda \in \mathbb{R}$ and $\vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu(-\hat{i} + 3\hat{j} + 2\hat{k}), \mu \in \mathbb{R}$ is

Options:

(A) $\sqrt{86}$

(B) $\sqrt{20}$

(C) $\sqrt{54}$

(D) $\sqrt{74}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You have a point $Q(0, 2, -2)$ floating in front of you and a point $P(5, -4, 3)$ defining a line.

We need to find the shortest distance from $Q$ to a line passing through $P$. The line is constrained to be perpendicular to two other lines, which is a classic JEE Advanced geometry challenge.

### Phase 1: The Cross Product—Finding the Direction

To define our line, we need a direction vector. We are given two reference lines with direction vectors $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 5\hat{k}$ and $\vec{b}_2 = -\hat{i} + 3\hat{j} + 2\hat{k}$.

Since our mystery line is perpendicular to both, its direction vector $\vec{b}$ must be the cross product of $\vec{b}_1$ and $\vec{b}_2$. We set up the determinant:

$$
\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 5 \ -1 & 3 & 2 \end{vmatrix}
$$

Expanding this, we get $\vec{b} = \hat{i}(6 - 15) - \hat{j}(4 - (-5)) + \hat{k}(6 - (-3))$, which simplifies to $\vec{b} = -9\hat{i} - 9\hat{j} + 9\hat{k}$.

We do not need the magnitude of this vector, only its direction. By dividing by $-9$, we obtain the much cleaner direction vector $\vec{d} = \hat{i} + \hat{j} - \hat{k}$.

### Phase 2: The Parametric Dance

Now that we have a point $P(5, -4, 3)$ and a direction $\vec{d} = \hat{i} + \hat{j} - \hat{k}$, we can write the equation of our line in parametric form: $\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})$.

Every point on this line is now accessible via the parameter $\lambda$. Let the foot of the perpendicular from $Q$ to this line be $M$.

The coordinates of $M$ are simply $(5+\lambda, \lambda-4, 3-\lambda)$.

### Phase 3: The Perpendicularity Condition

We need the vector $\vec{QM}$, which represents the path from $Q(0, 2, -2)$ to $M(5+\lambda, \lambda-4, 3-\lambda)$. Calculating this, we get $\vec{QM} = (5+\lambda)\hat{i} + (\lambda-6)\hat{j} + (5-\lambda)\hat{k}$.

Because $QM$ is the perpendicular distance, the vector $\vec{QM}$ must be perpendicular to the line's direction vector $\vec{d}$. This means their dot product must be zero:

$$
\vec{QM} \cdot (\hat{i} + \hat{j} - \hat{k}) = 0
$$

$$
(5+\lambda)(1) + (\lambda-6)(1) + (5-\lambda)(-1) = 0
$$

Solving this linear equation: $5 + \lambda + \lambda - 6 - 5 + \lambda = 0$, which simplifies to $3\lambda - 6 = 0$. This gives us the parameter value $\lambda = 2$.

### The Final Destination

With $\lambda = 2$, we find the specific vector $\vec{QM} = (5+2)\hat{i} + (2-6)\hat{j} + (5-2)\hat{k} = 7\hat{i} - 4\hat{j} + 3\hat{k}$.

The shortest distance is the magnitude of this vector:

$$
d = |\vec{QM}| = \sqrt{7^2 + (-4)^2 + 3^2} = \sqrt{49 + 16 + 9} = \sqrt{74}
$$

The shortest distance is **$\sqrt{74}$**. You have successfully navigated the 3D landscape; every complex problem is just a series of logical steps waiting for you to connect them.

Select Answer:

Visualized Solution

Visualizing the Problem

Extracting Direction Vectors

Cross Product Setup

Computing the Cross Product

Simplifying the Direction Vector

Equation of the Main Line

Defining the Foot of Perpendicular

Formulating Vector $QM$

Applying the Perpendicularity Condition

Solving for Parameter $λ$

Evaluating Vector $QM$

Final Distance Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Destination

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