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GUJCET · Maths · Three Dimensional Geometry
The vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines \(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) and \(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\) is ________.
- A \(\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} - 6\hat{k})\)
- B \(\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} - 3\hat{j} + 6\hat{k})\)
- C \(\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})\)
- D \(\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} - 3\hat{j} - 6\hat{k})\)
Answer & Solution
Correct Answer
(C) \(\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})\)
Step-by-step Solution
Detailed explanation
\(\vec{d_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}\), \(\vec{d_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}\) \(\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}\)
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