Question

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the XY and YZ planes at A and B. The line through A and B is}

• A. $[\vec{r}-(\hat{i}-2\hat{j}+0\hat{k})]\times(-\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{2}\hat{k})=\vec{0}$
• B. $[\vec{r}+(\hat{i}-2\hat{j}+0\hat{k})]\times(-\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k})=\vec{0}$
• C. $\vec{r}=(-\hat{i}-2\hat{j}+0\hat{k})+\lambda(-\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{2}\hat{k})$
• D. $\vec{r}=(\hat{i}+2\hat{j})+\lambda(-\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{2}\hat{k})$

Hint:

Points where a line meets coordinate planes are called the traces of the line.

Answer 1

The Correct Option is A

Step 1: Concept
A point is in the XY plane if $z=0$, and in the YZ plane if $x=0$.

Step 2: Analysis
- For XY plane ($z=0$): From the line equation, $\frac{x-1}{2} = \frac{y+2}{-1} = 0$. So $A = (1, -2, 0)$.
- For YZ plane ($x=0$): From the line equation, $\frac{0-1}{2} = \frac{y+2}{-1} = \frac{z}{1}$. So $z = -1/2$ and $y = -3/2$. $B = (0, -3/2, -1/2)$.

Step 3: Calculation
Direction vector $\vec{AB} = B - A = (-1, 1/2, -1/2)$.
The line passes through $A(1, -2, 0)$ with direction $(-1, 1/2, -1/2)$.
Vector equation: $(\vec{r} - \vec{a}) \times \vec{d} = 0$.

Step 4: Conclusion
Hence, the correct vector equation is given in option (A). Final Answer: (A)