Which of the following is not true? Given ove | vedclass

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(B) First,calculate the magnitudes of the vectors:$A = |\overrightarrow A| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.$B = |\overrightarrow B

Vedclass · Accepted Answer

$\frac{A}{B} = \frac{1}{4}$

Vedclass · Answer

(B) First,calculate the magnitudes of the vectors:$A = |\overrightarrow A| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.$B = |\overrightarrow B| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.Now,evaluate each option:$A$: $\overrightarrow A 	imes \overrightarrow B = (3\hat i + 4\hat j) 	imes (6\hat i + 8\hat j) = (3 	imes 8 - 4 	imes 6)\hat k = (24 - 24)\hat k = 0$. This is true.$B$: $\frac{A}{B} = \frac{5}{10} = \frac{1}{2}$. The statement $\frac{A}{B} = \frac{1}{4}$ is false.$C$: $\overrightarrow A \cdot \overrightarrow B = (3)(6) + (4)(8) = 18 + 32 = 50$. This is true.$D$: $A = 5$. This is true.Therefore,the statement that is not true is $\frac{A}{B} = \frac{1}{4}$.

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

Which of the following is not true? Given $\overrightarrow A = 3\hat i + 4\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$,where $A$ and $B$ are the magnitudes of $\overrightarrow A$ and $\overrightarrow B$.

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