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

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(D) Given $\vec{d} = \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}$.Since $\vec{a} \cdot \vec{d} = 1$,we have $\vec{a} \cdot \frac{\vec{b}+\vec{c}}{|\vec{

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$11$

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(D) Given $\vec{d} = \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}$.Since $\vec{a} \cdot \vec{d} = 1$,we have $\vec{a} \cdot \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|} = 1$,which implies $\vec{a} \cdot (\vec{b}+\vec{c}) = |\vec{b}+\vec{c}|$.Calculate $\vec{a} \cdot (\vec{b}+\vec{c}) = (\hat{i}+\hat{j}+\hat{k}) \cdot ((x+2)\hat{i} + 6\hat{j} - 2\hat{k}) = x+2+6-2 = x+6$.Calculate $|\vec{b}+\vec{c}| = |(x+2)\hat{i} + 6\hat{j} - 2\hat{k}| = \sqrt{(x+2)^2 + 6^2 + (-2)^2} = \sqrt{x^2+4x+4+36+4} = \sqrt{x^2+4x+44}$.Equating the two: $x+6 = \sqrt{x^2+4x+44}$.Squaring both sides: $(x+6)^2 = x^2+4x+44 \implies x^2+12x+36 = x^2+4x+44$.$8x = 8 \implies x = 1$.Now,calculate the scalar triple product $(\vec{a} 	imes \vec{b}) \cdot \vec{c} = [\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \ 2 & 4 & -5 \ x & 2 & 3 \end{vmatrix}$.Substituting $x=1$: $\begin{vmatrix} 1 & 1 & 1 \ 2 & 4 & -5 \ 1 & 2 & 3 \end{vmatrix} = 1(12+10) - 1(6+5) + 1(4-4) = 22 - 11 + 0 = 11$.Thus,the value is $11$.

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}$,$x \in R$. If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$,then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to

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