If the line joining the points A(1,0,0) and B | vedclass

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(D) Given points $A(1,0,0)$ and $B(0,0,1)$.The direction ratios of the line joining points $A$ and $B$ are $(0-1, 0-0, 1-0)$,which is $(-1, 0, 1)$.Sin

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$\frac{\pi}{2}$

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(D) Given points $A(1,0,0)$ and $B(0,0,1)$.The direction ratios of the line joining points $A$ and $B$ are $(0-1, 0-0, 1-0)$,which is $(-1, 0, 1)$.Since this line is normal to the plane $\pi$,the normal vector to the plane $\pi$ is $\vec{n_1} = -\hat{i} + 0\hat{j} + \hat{k}$.The plane $\pi$ passes through point $A(1,0,0)$,so its equation is $-1(x-1) + 0(y-0) + 1(z-0) = 0$,which simplifies to $-x + 1 + z = 0$ or $-x + z + 1 = 0$.The second plane is $x + y + z = 6$,so its normal vector is $\vec{n_2} = \hat{i} + \hat{j} + \hat{k}$.The angle $	heta$ between the two planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\vec{n_1} \cdot \vec{n_2} = (-1)(1) + (0)(1) + (1)(1) = -1 + 0 + 1 = 0$.Since the dot product is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

If the line joining the points $A(1,0,0)$ and $B(0,0,1)$ is a normal to the plane $\pi$ which passes through the point $A$,then the angle between the planes $\pi$ and $x+y+z=6$ is

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