If vec{a}, vec{b}, vec{c} are unit vectors su | vedclass

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \perp \vec{b}$ and $\vec{a} \p

Vedclass · Accepted Answer

$38$

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Since $\vec{a} \perp \vec{b}$ and $\vec{a} \perp \vec{c}$, we have $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$. The angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, so $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \frac{2 \pi}{3} = (1)(1) \left(-\frac{1}{2}\right) = -\frac{1}{2}$. Now, consider the expression $|\vec{a}+3 \vec{b}-4 \vec{c}|^2 = (\vec{a}+3 \vec{b}-4 \vec{c}) \cdot (\vec{a}+3 \vec{b}-4 \vec{c})$. Expanding this dot product: $|\vec{a}+3 \vec{b}-4 \vec{c}|^2 = |\vec{a}|^2 + 9|\vec{b}|^2 + 16|\vec{c}|^2 + 6(\vec{a} \cdot \vec{b}) - 8(\vec{a} \cdot \vec{c}) - 24(\vec{b} \cdot \vec{c})$. Substituting the known values: $|\vec{a}+3 \vec{b}-4 \vec{c}|^2 = 1 + 9(1) + 16(1) + 6(0) - 8(0) - 24\left(-\frac{1}{2}\right)$. $|\vec{a}+3 \vec{b}-4 \vec{c}|^2 = 1 + 9 + 16 + 12 = 38$.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $|\vec{a}+3 \vec{b}-4 \vec{c}|^2=$

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