vec{a}=2 hat{i}-hat{j}, vec{b}=2 hat{j}-hat{k | vedclass

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(D) Given $\vec{a}=2 \hat{i}-\hat{j}$, $\vec{b}=2 \hat{j}-\hat{k}$, $\vec{c}=-\hat{i}+2 \hat{k}$.Let $\vec{d} = x \hat{i} + y \hat{j} + z \hat{k}$ be

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$\sqrt{\frac{7}{2}}$

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(D) Given $\vec{a}=2 \hat{i}-\hat{j}$, $\vec{b}=2 \hat{j}-\hat{k}$, $\vec{c}=-\hat{i}+2 \hat{k}$.Let $\vec{d} = x \hat{i} + y \hat{j} + z \hat{k}$ be a unit vector such that $x^2+y^2+z^2=1$.Since $\vec{d} \perp \vec{c}$, we have $\vec{d} \cdot \vec{c} = 0 \Rightarrow -x + 2z = 0 \Rightarrow x = 2z$.Since $\vec{a}, \vec{b}, \vec{d}$ are coplanar, their scalar triple product is zero: $[\vec{a} \vec{b} \vec{d}] = 0$.$\begin{vmatrix} 2 & -1 & 0 \ 0 & 2 & -1 \ x & y & z \end{vmatrix} = 0$.Expanding the determinant: $2(2z + y) - (-1)(0 + x) = 0 \Rightarrow 4z + 2y + x = 0$.Substituting $x = 2z$: $4z + 2y + 2z = 0 \Rightarrow 2y = -6z \Rightarrow y = -3z$.Now, use the unit vector condition $x^2+y^2+z^2=1$: $(2z)^2 + (-3z)^2 + z^2 = 1 \Rightarrow 4z^2 + 9z^2 + z^2 = 1 \Rightarrow 14z^2 = 1 \Rightarrow z^2 = \frac{1}{14}$.We need to find $|\vec{d} \cdot \vec{b}| = |(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{j} - \hat{k})| = |2y - z|$.Substituting $y = -3z$: $|2(-3z) - z| = |-7z| = 7|z|$.Since $z^2 = \frac{1}{14}$, $|z| = \frac{1}{\sqrt{14}}$.Thus, $|\vec{d} \cdot \vec{b}| = 7 	imes \frac{1}{\sqrt{14}} = \frac{7}{\sqrt{14}} = \sqrt{\frac{49}{14}} = \sqrt{\frac{7}{2}}$.

$\vec{a}=2 \hat{i}-\hat{j}$, $\vec{b}=2 \hat{j}-\hat{k}$, $\vec{c}=2 \hat{k}-\hat{i}$ are three vectors and $\vec{d}$ is a unit vector perpendicular to $\vec{c}$. If $\vec{a}, \vec{b}, \vec{d}$ are coplanar vectors, then $|\vec{d} \cdot \vec{b}|=$

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