The vector(s) which is/are coplanar with vect | vedclass

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(A) Let $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$,$\vec{b} = \hat{i}+2\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+\hat{j}+\hat{k}$.The required vector $\vec{v}

Vedclass · Accepted Answer

$(A, D)$

Vedclass · Answer

(A) Let $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$,$\vec{b} = \hat{i}+2\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+\hat{j}+\hat{k}$.The required vector $\vec{v}$ is coplanar with $\vec{a}$ and $\vec{b}$,so $\vec{v} = x\vec{a} + y\vec{b}$.Also,$\vec{v}$ is perpendicular to $\vec{c}$,so $\vec{v} \cdot \vec{c} = 0$.$(x\vec{a} + y\vec{b}) \cdot \vec{c} = 0 \implies x(\vec{a} \cdot \vec{c}) + y(\vec{b} \cdot \vec{c}) = 0$.$\vec{a} \cdot \vec{c} = (1)(1) + (1)(1) + (2)(1) = 4$.$\vec{b} \cdot \vec{c} = (1)(1) + (2)(1) + (1)(1) = 4$.So,$4x + 4y = 0 \implies x = -y$.Thus,$\vec{v} = x(\vec{a} - \vec{b}) = x[(\hat{i}+\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}+\hat{k})] = x(-\hat{j}+\hat{k})$.For $x=1$,$\vec{v} = -\hat{j}+\hat{k}$ (Option $D$).For $x=-1$,$\vec{v} = \hat{j}-\hat{k}$ (Option $A$).Therefore,the vectors are $(A)$ and $(D)$.

The vector$(s)$ which is/are coplanar with vectors $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$,and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$ is/are:
$(A) \hat{j}-\hat{k}$
$(B) -\hat{i}+\hat{j}$
$(C) \hat{i}-\hat{j}$
$(D) -\hat{j}+\hat{k}$

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The vector$(s)$ which is/are coplanar with vectors $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$,and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$ is/are:$(A) \hat{j}-\hat{k}$$(B) -\hat{i}+\hat{j}$$(C) \hat{i}-\hat{j}$$(D) -\hat{j}+\hat{k}$