Fill in the blanks. The vector a + b bisects the angle between the non-collinear vectors a and b if _________.

The vector a + b bisects the angle between the non-collinear vectors a and b if a = b .Solution: If vector a = b bisects the angle between the non-collinear vectors, then a ( a + b )=| a | | a + b | b ( a + b )=b a^2+b^2 = a ( a +b) a a^2+b^2 ....(i) And b ( a + b )=| b | | a + b | [since, should be same] = b ( c +b) b a^2+b^2 ....(ii) From equations (i) and (ii), we get: a ( a +b) a a^2+b^2 = b ( c +b) b a^2+b^2 a | a | = b | b | a = b Thus, the vector a + b bisects the angle between the non-collinear vector a and b , if a and b are equal vectors.

Fill in the blanks. The vector a + b bisects the angle between th…

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Question

Fill in the blanks.
The vector $\vec{\text{a}}+\vec{\text{b}}$ bisects the angle between the non-collinear vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if _________.

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Answer

The vector $\vec{\text{a}}+\vec{\text{b}}$ bisects the angle between the non-collinear vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if $\vec{\text{a}}=\vec{\text{b}}.$Solution:
If vector $\vec{\text{a}}=\vec{\text{b}}$ bisects the angle between the non-collinear vectors, then $\vec{\text{a}}\cdot(\vec{\text{a}}+\vec{\text{b}})=|\vec{\text{a}}|\cdot|\vec{\text{a}}+\vec{\text{b}}|\cos\theta$ $\vec{\text{b}}\cdot(\vec{\text{a}}+\vec{\text{b}})=\text{b}\sqrt{\text{a}^2+\text{b}^2}\cos\theta$ $\Rightarrow\cos\theta=\frac{\vec{\text{a}}\cdot(\vec{\text{a}}+\vec{b})}{\text{a}\sqrt{\text{a}^2+\text{b}^2}}\ ....(\text{i})$ And $\vec{\text{b}}\cdot({\vec{\text{a}}}+\vec{\text{b}})=|\vec{\text{b}}|\cdot|\vec{\text{a}}+\vec{\text{b}}|\cos\theta$ [since, $\theta$ should be same] $\Rightarrow\cos\theta=\frac{\vec{\text{b}}\cdot(\vec{\text{c}}+\vec{b})}{\text{b}\sqrt{\text{a}^2+\text{b}^2}}\ ....(\text{ii})$ From equations (i) and (ii), we get: $\Rightarrow\frac{\vec{\text{a}}\cdot(\vec{\text{a}}+\vec{b})}{\text{a}\sqrt{\text{a}^2+\text{b}^2}}=\frac{\vec{\text{b}}\cdot(\vec{\text{c}}+\vec{b})}{\text{b}\sqrt{\text{a}^2+\text{b}^2}}$ $\Rightarrow\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{\vec{\text{b}}}{|\vec{\text{b}}|}$ $\therefore\vec{\text{a}}=\vec{\text{b}}$ Thus, the vector $\vec{\text{a}}+\vec{\text{b}}$ bisects the angle between the non-collinear vector $\vec{\text{a}}$ and $\vec{\text{b}},$ if $\vec{\text{a}}$ and $\vec{\text{b}}$ are equal vectors.