Let vec{a}=hat{i}+4 hat{j}+2 hat{k}, vec{b}=3 | vedclass

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(A) Given $\vec{d} 	imes \vec{b} = \vec{c} 	imes \vec{b}$.This implies $(\vec{d} - \vec{c}) 	imes \vec{b} = 0$.Therefore,$\vec{d} - \vec{c} = \lamb

Vedclass · Accepted Answer

$413$

Vedclass · Answer

(A) Given $\vec{d} 	imes \vec{b} = \vec{c} 	imes \vec{b}$.This implies $(\vec{d} - \vec{c}) 	imes \vec{b} = 0$.Therefore,$\vec{d} - \vec{c} = \lambda \vec{b}$,or $\vec{d} = \vec{c} + \lambda \vec{b}$ for some scalar $\lambda$.Given $\vec{d} \cdot \vec{a} = 24$,substitute $\vec{d} = \vec{c} + \lambda \vec{b}$ into this equation:$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24 \Rightarrow \vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 24$.Calculate the dot products:$\vec{a} \cdot \vec{c} = (1)(2) + (4)(-1) + (2)(4) = 2 - 4 + 8 = 6$.$\vec{a} \cdot \vec{b} = (1)(3) + (4)(-2) + (2)(7) = 3 - 8 + 14 = 9$.Substitute these values: $6 + \lambda(9) = 24 \Rightarrow 9\lambda = 18 \Rightarrow \lambda = 2$.Now,find $\vec{d} = \vec{c} + 2\vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + 2(3\hat{i} - 2\hat{j} + 7\hat{k}) = (2+6)\hat{i} + (-1-4)\hat{j} + (4+14)\hat{k} = 8\hat{i} - 5\hat{j} + 18\hat{k}$.Finally,calculate $|\vec{d}|^2 = 8^2 + (-5)^2 + 18^2 = 64 + 25 + 324 = 413$.

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$. If a vector $\vec{d}$ satisfies $\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{d} \cdot \vec{a}=24$,then $|\vec{d}|^2$ is equal to $.........$.

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