Let a straight line L pass through the point P(2,-1,3) and be perpendicular to the lines (x-1)/2 = (

Q: Why do we use the cross product here? Can't we just use dot products?

You absolutely could use dot products! If the direction vector of $L$ is $\vec{d} = (a, b, c)$, then $\vec{d} \cdot \vec{d}_1 = 0$ and $\vec{d} \cdot \vec{d}_2 = 0$. This would give you a system of two linear equations with three variables, which you could solve to find the ratios of $a, b,$ and $c$. However, the cross product is the direct, elegant tool specifically designed for this purpose—it gives you the orthogonal vector in one go!

Q: Why is the $x$-coordinate of $Q$ zero?

Think of the $yz$-plane as a flat sheet of paper standing vertically in 3D space. Any point on this plane has a height ($z$) and a depth ($y$), but it has no 'thickness' or 'width' along the $x$-axis. Therefore, every point on the $yz$-plane is defined by the equation $x = 0$. It is the fundamental constraint of that plane.

Q: Does it matter if I scale the direction vector?

Not at all! A line is defined by its direction, not the magnitude of its direction vector. If your cross product gives you $(10, -10, 5)$, you can divide by $5$ to get $(2, -2, 1)$. Both vectors point in the exact same direction, and using the smaller, simplified integers makes your algebra much cleaner and less prone to calculation errors.

Question

Let a straight line L pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$ and $\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}$ If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is

Options:

(A) $\sqrt{10}$

(B) $2\sqrt{3}$

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a point $P(2, -1, 3)$ and a line $L$ passing through $P$. This line $L$ is constrained to be perpendicular to two given lines, $L_1$ and $L_2$. Our goal is to find the point $Q$ where $L$ intersects the $yz$-plane and to calculate the distance $PQ$.

### Extracting the DNA of the Lines

Every line in 3D space carries its orientation in its direction vector. The symmetric equations for our lines are:

$$L_1: \frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-2}$$

$$L_2: \frac{x-3}{1} = \frac{y-2}{3} = \frac{z+2}{4}$$

From these, we extract the direction vectors $\vec{d}_1 = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d}_2 = \hat{i} + 3\hat{j} + 4\hat{k}$. These vectors represent the orientation of the lines in space.

### The Power of the Cross Product

We require a direction vector $\vec{d}$ for line $L$ that is perpendicular to both $\vec{d}_1$ and $\vec{d}_2$. We calculate this using the cross product $\vec{d} = \vec{d}_1 	imes \vec{d}_2$:

$$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 3 & 4 \end{vmatrix}$$

Expanding the determinant, we obtain:

$$\vec{d} = \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = 10\hat{i} - 10\hat{j} + 5\hat{k}$$

To simplify our calculations, we scale this vector by dividing by $5$, yielding the direction ratios $(2, -2, 1)$.

### Defining the Path

With point $P(2, -1, 3)$ and direction vector $(2, -2, 1)$, the symmetric equation of line $L$ is:

$$\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = t$$

By introducing the parameter $t$, any point $Q$ on line $L$ can be expressed as:

$$Q(2t+2, -2t-1, t+3)$$

### The Intersection

The point $Q$ lies on the $yz$-plane, which implies that its $x$-coordinate must be zero. Setting the $x$-component of $Q$ to zero:

$$2t + 2 = 0 \Rightarrow t = -1$$

Substituting $t = -1$ into the expressions for $y$ and $z$, we find:

$$y = -2(-1) - 1 = 1$$

$$z = -1 + 3 = 2$$

Thus, the intersection point is **$Q(0, 1, 2)$**.

### The Final Distance

We calculate the distance between $P(2, -1, 3)$ and $Q(0, 1, 2)$ using the distance formula:

$$PQ = \sqrt{(0 - 2)^2 + (1 - (-1))^2 + (2 - 3)^2}$$

$$PQ = \sqrt{(-2)^2 + (2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9}$$

The final distance is **$3$ units**.

Select Answer:

Visualized Solution

Visualizing the Problem Setup

Extracting Direction Vectors $d_{1}, d_{2}$

Cross Product $d = d_{1} \times d_{2}$

Evaluating the Determinant

Simplifying Direction Ratios

Equation of Line $L$

Parametric Coordinates of $Q$

Condition for $y z$ -plane

Solving for $t$

Exact Coordinates of $Q$

Distance Formula Setup

Final Distance Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Extracting the DNA of the Lines

The Power of the Cross Product

Defining the Path

The Intersection

The Final Distance

Similar Questions

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Let the plane $P : 8 x + α_{1} y + α_{2} z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ . If the intercept of $P$ on the $y$ -axis is 1, then the distance between $P$ and $L$ is :

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Select Answer:

Visualized Solution

Visualizing the Problem Setup

Extracting Direction Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​,d2​

Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=d1​×d2​

Evaluating the Determinant

Simplifying Direction Ratios

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yz-plane

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Exact Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q