Find the angle between the line r - i - 2j - k - - 2i - j - 2k and the plane r - 3i - 2j - 6k - 0-

Given, line is r=(i+2 j k)+μ (2i+j+2k) and the plane is r.(3i 2j+6k)=0 .Let θ be the angle between the line and the plane.Therefore sinθ = ...

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Angle between a Plane and a Line

Find the angl...

Question

Find the angle between the line $\to r = (\to i + 2 \to j - \to k) + μ (2 \to i + \to j + 2 \to k)$ and the plane $\to r . (3 \to i - 2 \to j + 6 \to k) = 0$ .

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\vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} - 2 \hat{k}) and \vec{r} = \hat{i} - \hat{j} + 2 \hat{k} - μ (2 \hat{i} + 4 \hat{j} - 4 \hat{k})

\vec{r} = (3 \hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 2 \hat{k}) and \vec{r} = (5 \hat{j} - 2 \hat{k}) + μ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})

\vec{r} = λ (\hat{i} + \hat{j} + 2 \hat{k}) and \vec{r} = 2 \hat{j} + μ \{(\sqrt{3} - 1) \hat{i} - (\sqrt{3} + 1) \hat{j} + 4 \hat{k}\}

\to r = 3 i + 2 j - 4 k + λ (i + 2 j + 2 k)

\to r = (5 j - 2 k) + μ (3 i + 2 j + 6 k)

\vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} - \hat{j} - \hat{k} + μ (2 \hat{i} + \hat{j} + 2 \hat{k})

\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}

\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) and \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k})

\vec{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}) and \vec{r} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k})

\to r = 3 i + 2 j - 4 k + λ (i + 2 j + 2 k)

\to r = 5 i - 2 k + μ (3 i + 2 j + 6 k)

\vec{r} = (2 \hat{i} - 5 \hat{j} + \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})

\vec{r} = 7 \hat{i} - 6 \hat{k} + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})

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