If vec a = 3vec j + 4vec k,vec b = 2vec i + v | vedclass

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(D) Given that $\vec c$ is the component of $\vec a$ parallel to $\vec b$ and $\vec d$ is the component of $\vec a$ perpendicular to $\vec b$. By defi

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$0$

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(D) Given that $\vec c$ is the component of $\vec a$ parallel to $\vec b$ and $\vec d$ is the component of $\vec a$ perpendicular to $\vec b$. By definition,$\vec a = \vec c + \vec d$. This implies that the vectors $\vec a$,$\vec c$,and $\vec d$ are coplanar. The scalar triple product of any three coplanar vectors is zero,i.e.,$[\vec a, \vec c, \vec d] = 0$. Let $\vec x = \vec a 	imes \vec c$,$\vec y = \vec c 	imes \vec d$,and $\vec z = \vec d 	imes \vec a$. The expression is $[\vec x 	imes \vec y, \vec y 	imes \vec z, \vec z 	imes \vec x]$. Using the identity for the scalar triple product of cross products,we have $[\vec x 	imes \vec y, \vec y 	imes \vec z, \vec z 	imes \vec x] = [\vec x, \vec y, \vec z]^2$. Since $\vec x, \vec y, \vec z$ are all perpendicular to the plane containing $\vec a, \vec c, \vec d$,they are parallel to each other. Thus,the vectors $\vec x, \vec y, \vec z$ are collinear,which implies $[\vec x, \vec y, \vec z] = 0$. Therefore,the value of the entire expression is $0^2 = 0$.

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Statement-$1$: If the points $(1, 2, 2), (2, 1, 2), (2, 2, z)$ and $(1, 1, 1)$ are coplanar,then $z = 2$.
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