Question

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Answer 1

2. If $\theta$ is the angle between $\vec{A}$ and $\vec{B}$ then the vector component of $\vec{B}$ acting along $\vec{A}$ will be,

$\longrightarrow\underline{\underline{\vec{B_A}=(B\cos\theta)\,\hat A}}$

3. So the vector component of $\vec{B}$ acting along $\vec{A}$ is,

$\longrightarrow\vec{B_A}=(B\cos\theta)\,\hat A$

$\longrightarrow\vec{B_A}=\left(\dfrac{AB\cos\theta}{A}\right)\,\hat A$

Since $AB\cos\theta=\vec{A}\cdot\vec{B}$ and $\hat A=\dfrac{\vec A}{A},$

$\longrightarrow\vec{B_A}=\dfrac{\vec{A}\cdot\vec{B}}{A}\cdot\dfrac{\vec A}{A}$

$\longrightarrow\vec{B_A}=\dfrac{\vec{A}\cdot\vec{B}}{A^2}\,\vec A$

If $\vec{A}=A_x\,\hat i+A_y\,\hat j,$

$\longrightarrow\vec{B_A}=\dfrac{\vec{A}\cdot\vec{B}}{(A_x)^2+(A_y)^2}\,(A_x\,\hat i+A_y\,\hat j)$

In the question,

• $\vec{A}=2\,\hat i+3\,\hat j$

• $\vec{B}=3\,\hat i+4\,\hat j$

Then, component of $\vec{B}$ acting along $\vec{A}$ is,

$\longrightarrow\vec{B_A}=\dfrac{(2\,\hat i+3\,\hat j)\cdot(3\,\hat i+4\,\hat j)}{2^2+3^2}\,(2\,\hat i+3\,\hat j)$

$\longrightarrow\vec{B_A}=\dfrac{6+12}{4+9}\,(2\,\hat i+3\,\hat j)$

$\longrightarrow\underline{\underline{\vec{B_A}=\dfrac{18}{13}\,(2\,\hat i+3\,\hat j)}}$

4. Here,

• $\vec A=3\,\hat i+4\,\hat j$

• $\vec B=2\,\hat i+3\,\hat j$

Then, component of $\vec{B}$ acting along $\vec{A}$ is,

$\longrightarrow\vec{B_A}=\dfrac{(3\,\hat i+4\,\hat j)\cdot(2\,\hat i+3\,\hat j)}{3^2+4^2}\,(3\,\hat i+4\,\hat j)$

$\longrightarrow\vec{B_A}=\dfrac{6+12}{9+16}\,(3\,\hat i+4\,\hat j)$

$\longrightarrow\underline{\underline{\vec{B_A}=\dfrac{18}{25}\,(3\,\hat i+4\,\hat j)}}$

Answer 2

Answer:answer is in the above attached sheet