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 \(6\hat{i}+2\hat{j}+3\hat{k}\) is:

• A. 3
• B. \(\frac{15}{7}\)
• C. \(\frac{18}{7}\)
• D. 4

Hint:

Finding the foot of a perpendicular from a point to a line is a standard procedure in 3D geometry.
1. Parameterize a general point on the line using \(\lambda\).
2. Form the vector from the given point to this general point.
3. Use the condition that this vector is perpendicular to the line's direction vector (dot product is zero).
4. Solve for \(\lambda\) and find the specific point.

Answer 1

The Correct Option is C

To determine the length of the projection of the vector \((\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k})\) on the vector \(6\hat{i}+2\hat{j}+3\hat{k}\), let's follow these steps:

First, identify the given line equation: 

\(\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})\).

The direction vector of the line is \(2\hat{i} + 3\hat{j} - \hat{k}\).

Given the point \((5, 4, 2)\), let \((\alpha, \beta, \gamma)\) be the foot of the perpendicular from this point to the line.

The vector equation from \((5, 4, 2)\) to \((\alpha, \beta, \gamma)\) can be expressed as:

\(\vec{AP} = (\alpha - 5)\hat{i} + (\beta - 4)\hat{j} + (\gamma - 2)\hat{k}\).

This vector \(\vec{AP}\) is perpendicular to the direction vector \(2\hat{i} + 3\hat{j} - \hat{k}\), hence:

1. \((\alpha - 5) \cdot 2 + (\beta - 4) \cdot 3 + (\gamma - 2)(-1) = 0\)

Solving, we get:

1. \(2(\alpha - 5) + 3(\beta - 4) - (\gamma - 2) = 0\) \(2\alpha + 3\beta - \gamma = 28\)

For \((\alpha, \beta, \gamma)\) to lie on the line, it should also satisfy:

1. \(\alpha = -1 + 2\lambda, \quad \beta = 3 + 3\lambda, \quad \gamma = 1 - \lambda\)

Substitute these parametric expressions into the equation \(2\alpha + 3\beta - \gamma = 28\):

• \(2(-1 + 2\lambda) + 3(3 + 3\lambda) - (1 - \lambda) = 28\)
• \(-2 + 4\lambda + 9 + 9\lambda - 1 + \lambda = 28\)
• \(14\lambda + 6 = 28\)
• \(14\lambda = 22 \implies \lambda = \frac{22}{14} = \frac{11}{7}\)

Substituting \(\lambda = \frac{11}{7}\) back into the expressions for \(\alpha, \beta, \gamma\):

• \(\alpha = -1 + 2\left(\frac{11}{7}\right) = -1 + \frac{22}{7} = \frac{15}{7}\)
• \(\beta = 3 + 3\left(\frac{11}{7}\right) = 3 + \frac{33}{7} = \frac{54}{7}\)
• \(\gamma = 1 - \left(\frac{11}{7}\right) = \frac{-4}{7}\)

The required projection length of \(\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}\) on \(6\hat{i} + 2\hat{j} + 3\hat{k}\) is:

\(\frac{(\alpha \cdot 6 + \beta \cdot 2 + \gamma \cdot 3)}{\sqrt{6^2 + 2^2 + 3^2}}\)

Substitute the values:

• \(L = \frac{\left(\frac{15}{7}\right) \times 6 + \left(\frac{54}{7}\right) \times 2 + \left(\frac{-4}{7}\right) \times 3}{\sqrt{49}} = \frac{52}{7}\cdot\frac{1}{7} = \frac{18}{7}\)

Thus, the length of the projection is \(\frac{18}{7}\), which matches the given correct answer.