Question
A vector $\vec{A}$ when added to the vector $\vec{B}=3 \hat{i}+4 \hat{j}$ yields a resultant vector that is in the positive $y$ -direction and has a magnitude equal to that of $\vec{B} .$ Find the magnitude of $\vec{A}$.a. $\sqrt{10}$b. 10c. 5d. $\sqrt{15}$
Step 1
The magnitude of a vector is given by the square root of the sum of the squares of its components. So, the magnitude of $\vec{B}$ is $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Show more…
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If $\vec{b}=3 \hat{i}+4 \hat{j}$ and $\vec{a}=\hat{i}-\hat{j}$, the vector having the same magnitude as that of $\vec{b}$ and parallel to $\vec{a}$ is a. $\frac{5}{\sqrt{2}}(\hat{i}-\hat{j})$. b. $\frac{5}{\sqrt{2}}(\hat{i}+\hat{j})$ c. $5(\hat{i}-\hat{j})$ d. $5(\hat{i}+\hat{j})$
If $\vec{b}=3 \hat{i}+4 \hat{j}$ and $\vec{a}=\hat{i}-\hat{j}$, the vector having the same magnitude as that of $\vec{b}$ and parallel to $\vec{a}$ is (A) $\frac{5}{\sqrt{2}}(\hat{i}-\hat{j})$ (B) $\frac{5}{\sqrt{2}}(\hat{i}+\hat{j})$ (C) $5(\hat{i}-\hat{j})$ (D) $5(\hat{i}+\hat{j})$
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