Competency-Based
Assessment Made Simple.

Step 1: Find the vector $\overrightarrow{BC}$

The position vectors of B and C are given as $\hat{i}+2\hat{j}-\hat{k}$ and $\hat{i}+5\hat{j}+3\hat{k}$ respectively. Therefore, the vector $\overrightarrow{BC}$ is given by the difference of their position vectors:

$\overrightarrow{BC} = (\hat{i}+5\hat{j}+3\hat{k}) - (\hat{i}+2\hat{j}-\hat{k}) = (1-1)\hat{i} + (5-2)\hat{j} + (3-(-1))\hat{k} = 0\hat{i} + 3\hat{j} + 4\hat{k} = 3\hat{j} + 4\hat{k}$

Step 2: Find the vector equation of the line passing through A and parallel to $\overrightarrow{BC}$

The position vector of point A is given as $3\hat{i}-\hat{j}-2\hat{k}$. The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\lambda$ is a scalar.

In this case, $\vec{a} = 3\hat{i}-\hat{j}-2\hat{k}$ and $\vec{b} = 3\hat{j} + 4\hat{k}$. Therefore, the vector equation of the line is:

$\vec{r} = (3\hat{i}-\hat{j}-2\hat{k}) + \lambda(3\hat{j} + 4\hat{k})$

Step 3: Find the cartesian equation of the line

The cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $a, b, c$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

Here, the point is A with position vector $3\hat{i}-\hat{j}-2\hat{k}$, so $(x_1, y_1, z_1) = (3, -1, -2)$. The direction ratios are given by the components of the vector $\overrightarrow{BC}$, which are $0, 3, 4$. Therefore, the cartesian equation of the line is:

$\frac{x-3}{0} = \frac{y-(-1)}{3} = \frac{z-(-2)}{4}$

$\frac{x-3}{0} = \frac{y+1}{3} = \frac{z+2}{4}$