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Question

Physics Question on Vector basics

Of the vectors given below, the parallel vectors are, A=6i^+8j^{{A}^{\to }}=6\widehat{i}+8\widehat{j} B=210i^+280k^{{B}^{\to }}=210\widehat{i}+280\widehat{k} C=5.1i^+6.8k^{{C}^{\to }}=5.1\widehat{i}+6.8\widehat{k} D=3.6i^+6j^+48k^{{D}^{\to }}=3.6\widehat{i}+6\widehat{j}+48\widehat{k}

A

A{{A}^{\to }} and B{{B}^{\to }}

B

A{{A}^{\to }} and C{{C}^{\to }}

C

A{{A}^{\to }} and D{{D}^{\to }}

D

C{{C}^{\to }} and D{{D}^{\to }}

Answer

A{{A}^{\to }} and D{{D}^{\to }}

Explanation

Solution

A=6i^+8j^\vec{A}=6\hat{i}+8\hat{j} B=210i^+280k^\vec{B}=210\hat{i}+280\hat{k} C=5.1i^+6.8j^=120(6i^+8j^)\vec{C}=5.1\hat{i}+6.8\hat{j}=\frac{1}{20}(6\hat{i}+8\hat{j}) D=3.6i^+8j^+4.8k^\vec{D}=3.6\hat{i}+8\hat{j}+4.8\hat{k} Hence, it is clear that \text{\vec{A}} and C\vec{C} are parallel and we can write as A=120C\vec{A}=\frac{1}{20}\vec{C} This implies that \text{\vec{A}} is parallel to C\vec{C} and magnitude of \text{\vec{A}} is 120.\frac{1}{20}.