If the vector 19 hat{i}+22 hat{j}+5 hat{k} bi | vedclass

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(B) Let vector $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Since the vector $v = 19 \hat{i} + 22 \hat{j} + 5 \hat{k}$ bisects the angle between $a$

Vedclass · Accepted Answer

$\frac{1}{3}(2 \hat{i}+2 \hat{j}+\hat{k})$

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(B) Let vector $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Since the vector $v = 19 \hat{i} + 22 \hat{j} + 5 \hat{k}$ bisects the angle between $a$ and $b = 6 \hat{i} + 8 \hat{j}$, it must be proportional to the sum of their unit vectors:$\lambda \left( \frac{a}{|a|} + \frac{b}{|b|} \right) = v$Given $|b| = \sqrt{6^2 + 8^2} = 10$, so $\frac{b}{|b|} = \frac{6 \hat{i} + 8 \hat{j}}{10} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$.Thus, $\frac{a}{|a|} = \frac{19 \hat{i} + 22 \hat{j} + 5 \hat{k}}{\lambda} - (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) = (\frac{19}{\lambda} - \frac{3}{5}) \hat{i} + (\frac{22}{\lambda} - \frac{4}{5}) \hat{j} + \frac{5}{\lambda} \hat{k}$.Since $\frac{a}{|a|}$ is a unit vector, its magnitude squared is $1$:$(\frac{19}{\lambda} - \frac{3}{5})^2 + (\frac{22}{\lambda} - \frac{4}{5})^2 + (\frac{5}{\lambda})^2 = 1$.Expanding this: $\frac{361}{\lambda^2} - \frac{114}{5\lambda} + \frac{9}{25} + \frac{484}{\lambda^2} - \frac{176}{5\lambda} + \frac{16}{25} + \frac{25}{\lambda^2} = 1$.$\frac{870}{\lambda^2} - \frac{290}{5\lambda} + 1 = 1 \Rightarrow \frac{870}{\lambda^2} = \frac{58}{\lambda} \Rightarrow \lambda = \frac{870}{58} = 15$.Substituting $\lambda = 15$ back, we get $\frac{a}{|a|} = (\frac{19}{15} - \frac{9}{15}) \hat{i} + (\frac{22}{15} - \frac{12}{15}) \hat{j} + \frac{5}{15} \hat{k} = \frac{10}{15} \hat{i} + \frac{10}{15} \hat{j} + \frac{5}{15} \hat{k} = \frac{1}{3}(2 \hat{i} + 2 \hat{j} + \hat{k})$.Therefore, the correct option is $(b)$.

If the vector $19 \hat{i}+22 \hat{j}+5 \hat{k}$ bisects an angle between the vectors $a$ and $6 \hat{i}+8 \hat{j}$, then the unit vector in the direction of $a$ is

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