Question

The point of intersection of the straight lines $\vec{r} = (3\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(-\hat{i} - 2\hat{j} + 2\hat{k})$ and $\frac{3-x}{-1} = \frac{y+4}{2} = \frac{z-5}{7}$ is:

• A. $(-3, -4, -5)$
• B. $(-3, 4, 5)$
• C. $(-3, 4, -5)$
• D. $(-3, -4, 5)$
• E. $(3, -4, 5)$

Hint:

When given options, you can bypass the algebra. Notice that the first line starts at $(3, -4, 5)$. Plug this point into the second line: $(3-3)/1 = 0$, $(-4+4)/2 = 0$, and $(5-5)/7 = 0$. Since $0=0=0$, the point is on both lines.

Answer 1

Answer: (E)

Concept: The point of intersection must satisfy the equations of both lines. To find it, we can express the general point on the first line in terms of $\lambda$ and substitute it into the second line's equation, or simply check the provided options against both equations.

Step 1: Find the general point on the first line.
The vector equation gives:
$x = 3 - \lambda$, $y = -4 - 2\lambda$, $z = 5 + 2\lambda$.

Step 2: Standardize the second line and substitute.
Line 2: $\frac{-(x-3)}{-1} = \frac{y+4}{2} = \frac{z-5}{7} \Rightarrow \frac{x-3}{1} = \frac{y+4}{2} = \frac{z-5}{7}$.
Substituting the expressions for $x, y, z$ from
Step 1:
$\frac{(3-\lambda)-3}{1} = \frac{(-4-2\lambda)+4}{2} = \frac{(5+2\lambda)-5}{7}$ $-\lambda = \frac{-2\lambda}{2} = \frac{2\lambda}{7}$

Step 3: Solve for $\lambda$.
The only value that satisfies $-\lambda = -\lambda = \frac{2\lambda}{7}$ is $\lambda = 0$.

Step 4: Determine the intersection point.
Using $\lambda = 0$ in the first line's parametric form: $x = 3 - 0 = 3$ $y = -4 - 0 = -4$ $z = 5 + 0 = 5$ Point $P = (3, -4, 5)$.