Question:medium
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:
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:
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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.
Updated On: May 2, 2026
- $(-3, -4, -5)$
- $(-3, 4, 5)$
- $(-3, 4, -5)$
- $(-3, -4, 5)$
- $(3, -4, 5)$
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The Correct Option is
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