Question

The straight line $\vec{r}=(2-3t)(\hat{i}+\hat{j})+(6t+1)(\hat{j}+\hat{k})+(12t-11)(\hat{k}+\hat{i}), t\in\mathbb{R}$, is parallel to the vector

• A. $3\hat{i}-\hat{j}+6\hat{k}$
• B. $3\hat{i}+\hat{j}+6\hat{k}$
• C. $3\hat{i}+\hat{j}-6\hat{k}$
• D. $-3\hat{i}+\hat{j}+6\hat{k}$
• E. $3\hat{i}+2\hat{j}+5\hat{k}$

Hint:

Math Tip: To find the direction ratios quickly, you only need to look at the coefficients of '$t$' in the original expression. The constants ($2, 1, -11$) determine the starting position but have absolutely no effect on the direction of the line!

Answer 1

The Correct Option is B

Concept:
Three-Dimensional Geometry - Vector Equation of a Line.
The standard vector equation of a line is $\vec{r} = \vec{a} + t\vec{d}$, where $\vec{a}$ is a position vector and $\vec{d}$ is the direction vector. The line is always parallel to its direction vector.
Step 1: Expand the given line equation.
Given: $\vec{r} = (2-3t)(\hat{i}+\hat{j}) + (6t+1)(\hat{j}+\hat{k}) + (12t-11)(\hat{k}+\hat{i})$ Distribute the scalar terms into their respective unit vectors: $$ \vec{r} = (2-3t)\hat{i} + (2-3t)\hat{j} + (6t+1)\hat{j} + (6t+1)\hat{k} + (12t-11)\hat{k} + (12t-11)\hat{i} $$
Step 2: Group by unit vector components ($\hat{i}, \hat{j}, \hat{k}$).
$$ \vec{r} = [(2-3t) + (12t-11)]\hat{i} + [(2-3t) + (6t+1)]\hat{j} + [(6t+1) + (12t-11)]\hat{k} $$
Step 3: Simplify the components.
Combine the constants and the '$t$' terms inside the brackets: $$ \vec{r} = (9t - 9)\hat{i} + (3t + 3)\hat{j} + (18t - 10)\hat{k} $$
Step 4: Separate into position and direction vectors.
Group the constants separately from the terms containing '$t$': $$ \vec{r} = (-9\hat{i} + 3\hat{j} - 10\hat{k}) + (9t\hat{i} + 3t\hat{j} + 18t\hat{k}) $$ Factor out '$t$' to reveal the standard form $\vec{r} = \vec{a} + t\vec{d}$: $$ \vec{r} = (-9\hat{i} + 3\hat{j} - 10\hat{k}) + t(9\hat{i} + 3\hat{j} + 18\hat{k}) $$
Step 5: Identify the parallel vector.
The direction vector is $\vec{d} = 9\hat{i} + 3\hat{j} + 18\hat{k}$.
The line is parallel to any vector that is a scalar multiple of $\vec{d}$.
Observe the options provided. By factoring out a $3$ from $\vec{d}$, we get: $$ \vec{d} = 3(3\hat{i} + \hat{j} + 6\hat{k}) $$ Therefore, the line is parallel to the vector $3\hat{i} + \hat{j} + 6\hat{k}$.