Let vec{a} = 3 hat{i} + 4 hat{j} - 5 hat{k} a | vedclass

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(A) The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} 	imes \vec{b}$.The unit vector perpendicu

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(A) The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} 	imes \vec{b}$.The unit vector perpendicular to the plane is $\hat{n} = \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|}$.The projection of the vector $(\vec{a} + \vec{b})$ on the vector $\hat{n}$ is given by $(\vec{a} + \vec{b}) \cdot \hat{n}$.Substituting $\hat{n}$,we get the projection as $(\vec{a} + \vec{b}) \cdot \frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \frac{(\vec{a} + \vec{b}) \cdot (\vec{a} 	imes \vec{b})}{|\vec{a} 	imes \vec{b}|}$.Using the properties of the scalar triple product,we have $(\vec{a} + \vec{b}) \cdot (\vec{a} 	imes \vec{b}) = \vec{a} \cdot (\vec{a} 	imes \vec{b}) + \vec{b} \cdot (\vec{a} 	imes \vec{b})$.Since the scalar triple product of vectors where two vectors are identical is zero,we have $[\vec{a}, \vec{a}, \vec{b}] = 0$ and $[\vec{b}, \vec{a}, \vec{b}] = 0$.Therefore,the projection is $\frac{0 + 0}{|\vec{a} 	imes \vec{b}|} = 0$.

Let $\vec{a} = 3 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 2 \hat{k}$. The projection of the sum of the vectors $\vec{a}$ and $\vec{b}$ on the vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is:

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