Find the length of the component of vec{a} al | vedclass

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Question

(A) The component of vector $\vec{a}$ along vector $\vec{b}$ is given by the projection formula: $	ext{Component} = \frac{\vec{a} \cdot \vec{b}}{|\ve

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) The component of vector $\vec{a}$ along vector $\vec{b}$ is given by the projection formula: $	ext{Component} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.First,calculate the dot product: $\vec{a} \cdot \vec{b} = (4 \hat{i} + 5 \hat{j}) \cdot (-\hat{i} + \hat{j}) = (4)(-1) + (5)(1) = -4 + 5 = 1$.Next,calculate the magnitude of vector $\vec{b}$: $|\vec{b}| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.Therefore,the length of the component is $\frac{1}{\sqrt{2}}$.

Column $I$	Column $II$
$(A)$ $\theta=60^{\circ}$	$(p)$ $n=\sqrt{3}$
$(B)$ $\theta=90^{\circ}$	$(q)$ $n=1$
$(C)$ $\theta=120^{\circ}$	$(r)$ $n=\sqrt{2}$
$(D)$ $\theta=180^{\circ}$	$(s)$ $n=2$

Find the length of the component of $\vec{a}$ along $\vec{b}$,where $\vec{a} = 4 \hat{i} + 5 \hat{j}$ and $\vec{b} = -\hat{i} + \hat{j}$.

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