Question

Let \( \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \), \( \vec{b} = 3\hat{i} + 3\hat{j} + \hat{k} \), and \( \vec{c} = \hat{i} - 2\hat{j} + 3\hat{k} \) be three vectors. If \( \vec{r} \) is a vector such that \( \vec{r} \times \vec{a} = \vec{r} \times \vec{b} \) and \( \vec{r} . \vec{c} = 18 \), then the magnitude of the orthogonal projection of \( 4\hat{i} + 3\hat{j} - \hat{k} \) on \( \vec{r} \) is:

• A. \( 4 \)
• B. \( 6 \)
• C. \( 12 \)
• D. \( 24 \)

Hint:

To find projection of one vector on another, use the formula: \[ \text{Magnitude of projection of } \vec{v} \text{ on } \vec{r} = \left| \frac{\vec{v} . \vec{r}}{|\vec{r}|} \right| \] If a vector is parallel to another, their cross product is zero.

Answer 1

The Correct Option is A

Step 1: Use the cross product condition
Given \( \vec{r} \times \vec{a} = \vec{r} \times \vec{b} \Rightarrow \vec{r} \times (\vec{a} - \vec{b}) = \vec{0} \). So \( \vec{r} \) is parallel to \( \vec{a} - \vec{b} \). \[ \vec{a} - \vec{b} = (2 - 3)\hat{i} + (1 - 3)\hat{j} + (3 - 1)\hat{k} = -\hat{i} - 2\hat{j} + 2\hat{k} \] Thus, \( \vec{r} = \lambda(-\hat{i} - 2\hat{j} + 2\hat{k}) \) for some scalar \( \lambda \). Step 2: Use dot product condition
Given \( \vec{r} . \vec{c} = 18 \), substitute \( \vec{r} \): \[ \lambda(-1\hat{i} -2\hat{j} + 2\hat{k}) . (\hat{i} - 2\hat{j} + 3\hat{k}) = 18 \] \[ \lambda[(-1)(1) + (-2)(-2) + (2)(3)] = 18 \Rightarrow \lambda[-1 + 4 + 6] = 18 \Rightarrow \lambda . 9 = 18 \Rightarrow \lambda = 2 \] So, \( \vec{r} = 2(-\hat{i} - 2\hat{j} + 2\hat{k}) = -2\hat{i} - 4\hat{j} + 4\hat{k} \) Step 3: Projection magnitude formula
We are asked to find the magnitude of the orthogonal projection of vector \( \vec{v} = 4\hat{i} + 3\hat{j} - \hat{k} \) on \( \vec{r} \): \[ \text{Projection magnitude} = \left| \frac{\vec{v} . \vec{r}}{|\vec{r}|} \right| \] Compute \( \vec{v} . \vec{r} \): \[ % Option (4)(-2) + (3)(-4) + (-1)(4) = -8 -12 -4 = -24 \] So \( |\vec{v} . \vec{r}| = 24 \) Compute \( |\vec{r}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \) \[ \text{Projection magnitude} = \left| \frac{-24}{6} \right| = 4 \]