Let (α, β, γ) be the co-ordinates of the foot of the perpendicular drawn from the point (5, 4, 2) on

Q: Why do we use the dot product to find the foot of the perpendicular?

The dot product is the most powerful tool for checking orthogonality. When two vectors are perpendicular, their dot product is zero. Since the segment PQ is perpendicular to the line, the vector PQ must be orthogonal to the line's direction vector. This is the most direct way to solve for the parameter \lambda.

Q: Can we use the distance formula instead of the dot product?

You could, but it would be much more tedious. Using the distance formula would involve minimizing the square of the distance, which leads to the same result but requires more algebraic heavy lifting. The dot product method is elegant and standard for JEE problems.

Q: What if the projection length comes out negative?

Remember that the length of a projection is a magnitude, which must always be non-negative. If your dot product calculation results in a negative value, you should take the absolute value of the dot product before dividing by the magnitude of the target vector.

Question

Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + \hat{2j} + 3\hat{k}$ is :

Options:

(A) $\frac{15}{7}$

(B) 4

(C) $\frac{18}{7}$

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the foot of the perpendicular $Q$ from point $P(5, 4, 2)$ to the line given by $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$, we first express a general point $Q$ on the line in terms of the parameter $\lambda$.

The coordinates of the general point $Q$ are:
$$Q = (-1 + 2\lambda, 3 + 3\lambda, 1 - \lambda)$$

### The Orthogonality Condition

We define the vector $\vec{PQ}$ by subtracting the coordinates of $P(5, 4, 2)$ from the coordinates of $Q$:
$$\vec{PQ} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-1 - \lambda)\hat{k}$$

For $Q$ to be the foot of the perpendicular, $\vec{PQ}$ must be orthogonal to the direction vector of the line, $\vec{d} = 2\hat{i} + 3\hat{j} - \hat{k}$. This implies their dot product must be zero:
$$\vec{PQ} \cdot \vec{d} = 0$$

Substituting the components into the dot product equation:
$$(2\lambda - 6)(2) + (3\lambda - 1)(3) + (-1 - \lambda)(-1) = 0$$

Expanding and simplifying the expression:
$$4\lambda - 12 + 9\lambda - 3 + 1 + \lambda = 0$$
$$14\lambda - 14 = 0$$
$$\lambda = 1$$

Substituting $\lambda = 1$ back into the general coordinates of $Q$, we find the foot of the perpendicular:
$$Q = (1, 6, 0)$$

### The Final Projection

We are tasked with finding the length of the projection of vector $\vec{a} = 1\hat{i} + 6\hat{j} + 0\hat{k}$ onto vector $\vec{b} = 6\hat{i} + 2\hat{j} + 3\hat{k}$. The formula for the projection length is given by:
$$	ext{Length} = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|}$$

First, calculate the dot product $\vec{a} \cdot \vec{b}$:
$$\vec{a} \cdot \vec{b} = (1)(6) + (6)(2) + (0)(3) = 6 + 12 + 0 = 18$$

Next, calculate the magnitude of vector $\vec{b}$:
$$|\vec{b}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$$

Thus, the final projection length is:
$$\mathbf{\frac{18}{7}}$$

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Defining a General Point on the Line

Constructing Vector $P Q$

The Condition for Perpendicularity

Solving for $λ$

Finding the Foot $(α, β, γ)$

Defining the Vectors for Projection

The Projection Formula

Calculating Dot Product and Magnitude

Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Orthogonality Condition

The Final Projection

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