A bisector of the angle between the normals o | vedclass

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(D) The normals to the given planes $4x + 3y - 5 = 0$ and $x + 2y + 2z - 4 = 0$ are $\vec{n_1} = 4\hat{i} + 3\hat{j} + 0\hat{k}$ and $\vec{n_2} = \hat

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$7\hat{i} - \hat{j} - 10\hat{k}$

Vedclass · Answer

(D) The normals to the given planes $4x + 3y - 5 = 0$ and $x + 2y + 2z - 4 = 0$ are $\vec{n_1} = 4\hat{i} + 3\hat{j} + 0\hat{k}$ and $\vec{n_2} = \hat{i} + 2\hat{j} + 2\hat{k}$.To find the bisector of the angle between these normals, we first normalize the vectors:$|\vec{n_1}| = \sqrt{4^2 + 3^2 + 0^2} = 5$$|\vec{n_2}| = \sqrt{1^2 + 2^2 + 2^2} = 3$The unit vectors are $\hat{n_1} = \frac{4}{5}\hat{i} + \frac{3}{5}\hat{j}$ and $\hat{n_2} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.The angle bisector is along the vector $\hat{n_1} \pm \hat{n_2}$.Taking the sum: $(\frac{4}{5} + \frac{1}{3})\hat{i} + (\frac{3}{5} + \frac{2}{3})\hat{j} + \frac{2}{3}\hat{k} = \frac{17}{15}\hat{i} + \frac{19}{15}\hat{j} + \frac{10}{15}\hat{k}$, which is proportional to $17\hat{i} + 19\hat{j} + 10\hat{k}$.Taking the difference: $(\frac{4}{5} - \frac{1}{3})\hat{i} + (\frac{3}{5} - \frac{2}{3})\hat{j} - \frac{2}{3}\hat{k} = \frac{7}{15}\hat{i} - \frac{1}{15}\hat{j} - \frac{10}{15}\hat{k}$, which is proportional to $7\hat{i} - \hat{j} - 10\hat{k}$.Comparing with the options, $7\hat{i} - \hat{j} - 10\hat{k}$ is the correct vector.

$A$ bisector of the angle between the normals of the planes $4x + 3y = 5$ and $x + 2y + 2z = 4$ is along the vector

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