In triangle OAB, O(0, 0, 0), A(6, 2, -3) and | vedclass

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(B) Given $\vec{a} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = 4\hat{i} + 0\hat{j} + 3\hat{k}$.Projection of $\vec{a}$ on $\vec{b}$ is $\vec{OM}

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$2\sqrt{10} 	ext{ units}$

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(B) Given $\vec{a} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = 4\hat{i} + 0\hat{j} + 3\hat{k}$.Projection of $\vec{a}$ on $\vec{b}$ is $\vec{OM} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} ight) \vec{b}$.$\vec{a} \cdot \vec{b} = (6)(4) + (2)(0) + (-3)(3) = 24 + 0 - 9 = 15$.$|\vec{b}|^2 = 4^2 + 0^2 + 3^2 = 16 + 9 = 25$.So, $\vec{OM} = \frac{15}{25} \vec{b} = \frac{3}{5} (4\hat{i} + 3\hat{k}) = \frac{12}{5}\hat{i} + \frac{9}{5}\hat{k}$.Vector $\vec{AM} = \vec{OM} - \vec{a} = (\frac{12}{5} - 6)\hat{i} + (0 - 2)\hat{j} + (\frac{9}{5} - (-3))\hat{k} = -\frac{18}{5}\hat{i} - 2\hat{j} + \frac{24}{5}\hat{k}$.$l(AM) = |\vec{AM}| = \sqrt{(-\frac{18}{5})^2 + (-2)^2 + (\frac{24}{5})^2} = \sqrt{\frac{324}{25} + 4 + \frac{576}{25}} = \sqrt{\frac{900}{25} + 4} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} 	ext{ units}$.

In $\triangle OAB$, $O(0, 0, 0)$, $A(6, 2, -3)$ and $B(4, 0, 3)$ are the vertices. Let $\vec{a}$ and $\vec{b}$ be position vectors of points $A$ and $B$ respectively. If $OM$ is the projection of $\vec{a}$ on $\vec{b}$, then the length $l(AM)$ is equal to...

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