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If `vecr=(hati+2hatj+3hatk)+lamda(hati-hatj+hatk) and vecr=(hati+2hatj+3hatk)+mu(hati+hatj-hatk)` are two lines, then find the equation of acute angle bisector of two lines. |
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Answer» Lines are `vecr=(hati+2hatj+3hatk)+lamda(hati-hatj+hatk) and vecr=(hati+2hatj+3hatk)+mu(hati+hatj-hatk)` along vectors `(hati-hatj+hatk) and (hati+hatj-hatk)`, respectively. `" "` Angle between two lines = `cos^(-1)""(((1)xx(1)+(-1)(1)+(1)(-1))/(sqrt(3)sqrt(3)))=cos^(-1)(-(1)/(3))` which is an obtuse angle. `therefore" "` Vector along acute angle bisector = `lamda[(hati-hatj+hatk)/(sqrt(3))-(hati+hatj-hatk)/(sqrt(3))]=(2lamda)/(sqrt(3))(-hatj+hatk)` `therefore" "` Equation of acute angle bisector = `(hati+2hatj+3hatk)+t(hatj-hatk)` |
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