Let two planes be P_1 : 2x - y + z = 2 and P_ | vedclass

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(B) The given planes are $P_1: 2x - y + z - 2 = 0$ and $P_2: x + 2y - z - 3 = 0$.The equation of the angle bisectors is given by $\frac{2x - y + z - 2

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$3x + y - 5 = 0$

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(B) The given planes are $P_1: 2x - y + z - 2 = 0$ and $P_2: x + 2y - z - 3 = 0$.The equation of the angle bisectors is given by $\frac{2x - y + z - 2}{\sqrt{2^2 + (-1)^2 + 1^2}} = \pm \frac{x + 2y - z - 3}{\sqrt{1^2 + 2^2 + (-1)^2}}$.Since the denominators are equal,we have $2x - y + z - 2 = \pm (x + 2y - z - 3)$.Case $1$ (Positive sign): $2x - y + z - 2 = x + 2y - z - 3 \implies x - 3y + 2z + 1 = 0$.Case $2$ (Negative sign): $2x - y + z - 2 = -x - 2y + z + 3 \implies 3x + y - 5 = 0$.To identify the acute angle bisector,check the sign of $a_1a_2 + b_1b_2 + c_1c_2$ where $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are coefficients of $x, y, z$ after making the constant terms positive. Rewriting planes as $P_1: -2x + y - z + 2 = 0$ and $P_2: x + 2y - z - 3 = 0$ is not standard; let's use $P_1: -2x + y - z + 2 = 0$ and $P_2: x + 2y - z - 3 = 0$. Actually,for $P_1: 2x - y + z - 2 = 0$ and $P_2: x + 2y - z - 3 = 0$,$a_1a_2 + b_1b_2 + c_1c_2 = (2)(1) + (-1)(2) + (1)(-1) = 2 - 2 - 1 = -1 Since the product is negative,the positive sign gives the obtuse angle bisector and the negative sign gives the acute angle bisector.Thus,the acute angle bisector is $3x + y - 5 = 0$.

Let two planes be $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Based on the given information,the equation of the acute angle bisector of the planes $P_1$ and $P_2$ is...

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