Saved Bookmarks
| 1. |
The acute angle between the line `r=(hati+2hatj+hatk)+lambda(hati+hatj+hatk)` and the plane `r.(2hati-hatj+hatk)=5`A. `cos^(-1)((sqrt2)/(3))`B. `sin^(-1)((sqrt2)/(3))`C. `tan^(-1)((sqrt2)/(3))`D. `sin^(-1)((sqrt2)/(3))` |
|
Answer» Correct Answer - B We know that, the angle between the line `r=a+lambdab` and the planer r.n=d is `sin theta =(n.b)/(|n||b|)` Given equation of line is `r=(hati+2hatj+hatk)+lambda(hati+hatj+hatk)` and plane is `r.(2hati-hatj+hatk)=5` Here, `b=hati+hatj+hatk and n=2hati-hatj+hatk` `therefore sin theta=((2hati-hatj+hatk).(hati+hatj+hatk))/(|2hati-hatj+hatk|hati+hatj+hatk))` `=(2-1+1)/(sqrt(2^2+(-1)^(2)+(1)^(2)sqrt(1^(2)+1^(2)+1^(2))` `(2)/(sqrt(4+1+1)sqrt(1+1+1))=(2)/(sqrt6 sqrt3))=(2)/(3sqrt2))` `Rightarrow sin theta=(sqrt2)/(3)` `Rightarrow theta=sin^(-1) ((sqrt2)/(3))` |
|