If the vectors 2 bar{i} + 4 bar{j} - 3 bar{k} | vedclass

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(D) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.Given vectors a

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$\frac{1}{9}(-7 \bar{i} - 4 \bar{j} + 4 \bar{k})$

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(D) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.Given vectors are $\vec{a} = 2 \bar{i} + 4 \bar{j} - 3 \bar{k}$,$\vec{b} = -\bar{i} + 2 \bar{j} + 3 \bar{k}$,and $\vec{c} = p \bar{i} - 2 \bar{j} + \bar{k}$.The condition $[\vec{a} \vec{b} \vec{c}] = 0$ implies the determinant of the components is zero:$\begin{vmatrix} 2 & 4 & -3 \ -1 & 2 & 3 \ p & -2 & 1 \end{vmatrix} = 0$Expanding along the first row:$2(2(1) - 3(-2)) - 4(-1(1) - 3(p)) - 3(-1(-2) - 2(p)) = 0$$2(2 + 6) - 4(-1 - 3p) - 3(2 - 2p) = 0$$2(8) + 4 + 12p - 6 + 6p = 0$$16 + 4 - 6 + 18p = 0$$14 + 18p = 0 \implies 18p = -14 \implies p = -\frac{7}{9}$.Now,substitute $p = -\frac{7}{9}$ into the vector $9p \bar{i} - 4 \bar{j} + 4 \bar{k}$:$9(-\frac{7}{9}) \bar{i} - 4 \bar{j} + 4 \bar{k} = -7 \bar{i} - 4 \bar{j} + 4 \bar{k}$.Let $\vec{v} = -7 \bar{i} - 4 \bar{j} + 4 \bar{k}$. The magnitude is $|\vec{v}| = \sqrt{(-7)^2 + (-4)^2 + 4^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$.The unit vector is $\frac{\vec{v}}{|\vec{v}|} = \frac{1}{9}(-7 \bar{i} - 4 \bar{j} + 4 \bar{k})$.

If the vectors $2 \bar{i} + 4 \bar{j} - 3 \bar{k}$,$-\bar{i} + 2 \bar{j} + 3 \bar{k}$,and $p \bar{i} - 2 \bar{j} + \bar{k}$ are coplanar,then the unit vector in the direction of the vector $9p \bar{i} - 4 \bar{j} + 4 \bar{k}$ is

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