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The angle between the planes `vecr.(2hati-hatj+hatk)=6` and `vecr.(hati+hatj+2hatk)=5` isA. `(pi)/3`B. `(2pi)/3`C. `(pi)/6`D. `(5pi)/6` |
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Answer» Correct Answer - A We know that the angle between the planes `vecr.vecn_(1)=d_(1)` and`vecr.vecn_(2)=d_(2)` is given by `cos theta=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)` Here `vecn_(1)=2hati-hatj+hatk` and `vecn_(2)=hati+hatj+2hatk` `:.cos theta=((2hati-hatj+hatk).(hati+hatj+2hatk))/(|2hati-hatj+hatk||hati+hatj+2hatk|)=1/2` `impliestheta=pi//3` If `a_(1)x+b_(1)y+c_(1)z+d_(1)=0` and `a_(2)x+b_(2)y+c_(2)z+d_(2)=0` are Cartesian equations of two planes, then vectors normal to them are `vecn_(1)=a_(1)hati+b_(1)hatj+c_(1)hatk` and `vecn_(2)=a_(2)hati+b_(2)hatj+c_(2)hatk` respectively. Therefore, the angle `theta` between the planes is given by `cos thet=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)` `impliescos theta=(a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2))sqrt(a_(1)^(2)+b_(2)^(2)+c_(2)^(2)))` If the planes are perpendicular then `vecn_(1)` and `vecn_(2)` are perpendicular. `:.vecn_(1).vecn_(2)=0impliesa_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0` If the planes are parallel then `vecn_(1)` and `vecn_(2)` are parallel. `:.(a_(1))/(a_(2))=(b_(1))/(b_(2))=(c_(1))/(c_(2))` |
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