If the given planes ax + by + cz + d = 0 and | vedclass

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(D) The normal vectors to the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are $\vec{n_1} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{

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$aa' + bb' + cc' = 0$

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(D) The normal vectors to the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are $\vec{n_1} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{n_2} = a'\hat{i} + b'\hat{j} + c'\hat{k}$ respectively.Two planes are mutually perpendicular if and only if their normal vectors are perpendicular.Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{n_1} \cdot \vec{n_2} = 0$.Calculating the dot product: $(a\hat{i} + b\hat{j} + c\hat{k}) \cdot (a'\hat{i} + b'\hat{j} + c'\hat{k}) = aa' + bb' + cc' = 0$.Therefore,the correct condition is $aa' + bb' + cc' = 0$.

If the given planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are mutually perpendicular,then:

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