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

Step 1: Understanding the Question:
The problem consists of two parts. First, we need to find the coordinates of the foot of the perpendicular from a given point P to a given line L. Let's call this foot F\((\alpha, \beta, \gamma)\). Second, we need to find the length of the projection of the position vector of F, \(\vec{OF}\), onto another given vector.
Step 2: Key Formula or Approach:
1. Foot of Perpendicular: Any point on the line \(\vec{r} = \vec{a} + \lambda\vec{d}\) can be written in coordinate form. Let this be the foot F. The vector connecting the given point P to F, i.e., \(\vec{PF}\), will be perpendicular to the direction vector of the line, \(\vec{d}\). Their dot product will be zero: \(\vec{PF} \cdot \vec{d} = 0\). This condition allows us to solve for \(\lambda\).
2. Projection of a Vector: The length of the projection of a vector \(\vec{v}\) onto a vector \(\vec{u}\) is given by \(\frac{|\vec{v} \cdot \vec{u}|}{|\vec{u}|}\).
Step 3: Detailed Explanation:
Part 1: Finding the foot of the perpendicular \(F(\alpha, \beta, \gamma)\).
The given point is P(5, 4, 2).
The given line is \(\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k})\). Any point on this line, which we assume is F, has coordinates: \[ F = (-1+2\lambda, 3+3\lambda, 1-\lambda) \] The direction vector of the line is \(\vec{d} = 2\hat{i}+3\hat{j}-\hat{k}\).
The vector \(\vec{PF}\) is given by:
\[ \vec{PF} = \vec{F} - \vec{P} = ((-1+2\lambda)-5)\hat{i} + ((3+3\lambda)-4)\hat{j} + ((1-\lambda)-2)\hat{k} \] \[ \vec{PF} = (2\lambda-6)\hat{i} + (3\lambda-1)\hat{j} + (-\lambda-1)\hat{k} \] Since \(\vec{PF}\) is perpendicular to \(\vec{d}\), their dot product is zero:
\[ \vec{PF} \cdot \vec{d} = 0 \] \[ (2\lambda-6)(2) + (3\lambda-1)(3) + (-\lambda-1)(-1) = 0 \] \[ 4\lambda - 12 + 9\lambda - 3 + \lambda + 1 = 0 \] \[ 14\lambda - 14 = 0 \implies \lambda = 1 \] Now substitute \(\lambda=1\) back into the coordinates of F: \[ \alpha = -1 + 2(1) = 1 \] \[ \beta = 3 + 3(1) = 6 \] \[ \gamma = 1 - 1 = 0 \] So, the foot of the perpendicular is F(1, 6, 0).
Part 2: Finding the length of the projection.
The vector from the origin to F is \(\vec{v} = \alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} = 1\hat{i}+6\hat{j}+0\hat{k}\).
We need to project this vector onto \(\vec{u} = 6\hat{i}+2\hat{j}+3\hat{k}\).
The length of the projection is \(\frac{|\vec{v} \cdot \vec{u}|}{|\vec{u}|}\).
First, calculate the dot product: \[ \vec{v} \cdot \vec{u} = (1)(6) + (6)(2) + (0)(3) = 6 + 12 + 0 = 18 \] Next, calculate the magnitude of \(\vec{u}\): \[ |\vec{u}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36+4+9} = \sqrt{49} = 7 \] The length of the projection is: \[ \text{Projection Length} = \frac{18}{7} \] Step 4: Final Answer:
The length of the projection is \(\frac{18}{7}\).