A unit vector vec{e} = a hat{i} + b hat{j} + | vedclass

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(B) Since $\vec{e}$ is coplanar with $\vec{u} = \hat{i} - 3 \hat{j} + 5 \hat{k}$ and $\vec{v} = 3 \hat{i} + \hat{j} - 5 \hat{k}$, the scalar triple pr

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$3$

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(B) Since $\vec{e}$ is coplanar with $\vec{u} = \hat{i} - 3 \hat{j} + 5 \hat{k}$ and $\vec{v} = 3 \hat{i} + \hat{j} - 5 \hat{k}$, the scalar triple product is zero:$\begin{vmatrix} a & b & c \ 1 & -3 & 5 \ 3 & 1 & -5 \end{vmatrix} = 0$$a(15 - 5) - b(-5 - 15) + c(1 + 9) = 0$$10a + 20b + 10c = 0 \Rightarrow a + 2b + c = 0$ ...$(i)$Given $\vec{e} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$, we have:$a + b + c = 0$ ...$(ii)$Subtracting $(ii)$ from $(i)$: $(a + 2b + c) - (a + b + c) = 0 \Rightarrow b = 0$.Substituting $b = 0$ into $(ii)$, $c = -a$.Since $\vec{e}$ is a unit vector, $a^2 + b^2 + c^2 = 1 \Rightarrow a^2 + 0^2 + (-a)^2 = 1 \Rightarrow 2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$ and $c^2 = \frac{1}{2}$.Now, $2a^2 + 3b^2 + 4c^2 = 2(\frac{1}{2}) + 3(0) + 4(\frac{1}{2}) = 1 + 0 + 2 = 3$.

$A$ unit vector $\vec{e} = a \hat{i} + b \hat{j} + c \hat{k}$ is coplanar with the vectors $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $3 \hat{i} + \hat{j} - 5 \hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$, then $2 a^2 + 3 b^2 + 4 c^2 =$

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