Let vec{a}, vec{b}, vec{c} be co-initial vect | vedclass

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(C) The vector along the angle bisector of two vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{c} = \lambda \left( \frac{\vec{a}}{|\vec{a}|} + \frac

Vedclass · Accepted Answer

$(2 x+3 y) \hat{i}+(7 y-x) \hat{j}+(5 x-y) \hat{k}$

Vedclass · Answer

(C) The vector along the angle bisector of two vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{c} = \lambda \left( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right)$.Given $\vec{a} = 2 \hat{i} - \hat{j} + 5 \hat{k}$,we have $|\vec{a}| = \sqrt{2^2 + (-1)^2 + 5^2} = \sqrt{4 + 1 + 25} = \sqrt{30}$.Given $\vec{b} = 3 \hat{i} + 7 \hat{j} - \hat{k}$,we have $|\vec{b}| = \sqrt{3^2 + 7^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.Let $x = \frac{1}{|\vec{a}|} = \frac{1}{\sqrt{30}}$ and $y = \frac{1}{|\vec{b}|} = \frac{1}{\sqrt{59}}$.Then $\vec{c} = \lambda (x \vec{a} + y \vec{b}) = \lambda (x(2 \hat{i} - \hat{j} + 5 \hat{k}) + y(3 \hat{i} + 7 \hat{j} - \hat{k}))$.Grouping the components,we get $\vec{c} = \lambda ((2x + 3y) \hat{i} + (7y - x) \hat{j} + (5x - y) \hat{k})$.Comparing this with the options,option $C$ represents the vector along the bisector.

Let $\vec{a}, \vec{b}, \vec{c}$ be co-initial vectors and $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=3 \hat{i}+7 \hat{j}-\hat{k}$. Let $(\vec{a}, \vec{b})=\theta$ be an acute angle and $\vec{c}$ be the vector along the bisector of the angle $\theta$. If $\lambda, x, y \in R$,then $\vec{c}=$

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