The vector that is parallel to the vector 2 h | vedclass

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(C) Let the required vector be $\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$.Since $\vec{v}$ is coplanar with $\vec{u_1} = \hat{i} + \hat{j}$ and $\ve

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$\hat{i} - \hat{j} - 2 \hat{k}$

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(C) Let the required vector be $\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$.Since $\vec{v}$ is coplanar with $\vec{u_1} = \hat{i} + \hat{j}$ and $\vec{u_2} = \hat{j} + \hat{k}$,it must be a linear combination of $\vec{u_1}$ and $\vec{u_2}$.Thus,$a \hat{i} + b \hat{j} + c \hat{k} = \lambda(\hat{i} + \hat{j}) + \mu(\hat{j} + \hat{k}) = \lambda \hat{i} + (\lambda + \mu) \hat{j} + \mu \hat{k}$.Comparing coefficients,we get $a = \lambda$,$c = \mu$,and $b = \lambda + \mu$. Substituting $\lambda$ and $\mu$,we get $b = a + c$.The vector $\vec{v}$ is parallel to $2 \hat{i} - 2 \hat{j} - 4 \hat{k}$,so $\vec{v} = k(2 \hat{i} - 2 \hat{j} - 4 \hat{k})$.For option $C$,$\vec{v} = \hat{i} - \hat{j} - 2 \hat{k}$,we have $a = 1, b = -1, c = -2$.Checking the condition $b = a + c$: $1 + (-2) = -1$,which matches $b$.Thus,the vector $\hat{i} - \hat{j} - 2 \hat{k}$ is the correct answer.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$,then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then,its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$,then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

The vector that is parallel to the vector $2 \hat{i} - 2 \hat{j} - 4 \hat{k}$ and coplanar with the vectors $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ is

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