If overrightarrow{A}=3 hat{imath}-2 hat{jmath | vedclass

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(D) Given vectors are $\vec{A}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k}$,$\vec{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k}$,and $\vec{C}=2 \hat{\imath}+\hat

Vedclass · Accepted Answer

$\overrightarrow{A}=\overrightarrow{B}+\overrightarrow{C}, B^{2}=A^{2}+C^{2}$

Vedclass · Answer

(D) Given vectors are $\vec{A}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k}$,$\vec{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k}$,and $\vec{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}$.First,check the vector addition relation:$\vec{B}+\vec{C} = (\hat{\imath}-3 \hat{\jmath}+5 \hat{k}) + (2 \hat{\imath}+\hat{\jmath}-4 \hat{k}) = 3 \hat{\imath}-2 \hat{\jmath}+\hat{k} = \vec{A}$.Thus,$\vec{A}=\vec{B}+\vec{C}$ is satisfied.Now,calculate the squares of the magnitudes:$A^2 = |\vec{A}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14$.$B^2 = |\vec{B}|^2 = 1^2 + (-3)^2 + 5^2 = 1 + 9 + 25 = 35$.$C^2 = |\vec{C}|^2 = 2^2 + 1^2 + (-4)^2 = 4 + 1 + 16 = 21$.Observing the values,$B^2 = 35$ and $A^2 + C^2 = 14 + 21 = 35$.Therefore,$B^2 = A^2 + C^2$ is satisfied.Hence,the correct option is $\vec{A}=\vec{B}+\vec{C}$ and $B^2=A^2+C^2$.

Column $I$	Column $II$
$(A)$ $\vec{A} \cdot \vec{B} = \|\vec{A} \times \vec{B}\|$	$(p)$ $\theta = 45^{\circ}$ or $135^{\circ}$
$(B)$ $\vec{A} \cdot \vec{B} = B^2$	$(q)$ $\theta = 0^{\circ}$
$(C)$ $\|\vec{A} + \vec{B}\| = \|\vec{A} - \vec{B}\|$	$(r)$ $\vec{A} = \vec{B}$
$(D)$ $\|\vec{A} \times \vec{B}\| = AB$	$(s)$ $\theta = 90^{\circ}$

Column $I$	Column $II$
$(A) \|A+B\|$	$(p) \frac{\sqrt{3}}{2} x^2$
$(B) \|A-B\|$	$(q) x$
$(C) A \cdot B$	$(r) \sqrt{3} x$
$(D) \|A \times B\|$	$(s) \frac{x^2}{2}$

If $\overrightarrow{A}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k}$,$\overrightarrow{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k}$ and $\overrightarrow{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}$ form a right-angled triangle,then which of the following is satisfied?

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