The distance of the point, (–1, –5, –10) from the point of intersection of the line, x-2…
Mathematics · JEE Main · NTA Exams — Three Dimensional Geometry
The distance of the point, (–1, –5, –10) from the point of intersection of the line, \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) and the plane, x – y + z = 5, is :
- 10
- 11
- 12
- 13
Answer
(D) 13
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- If the straight lines x = 1 + s, y = −3 − λ s, z = 1 + λs and x = t 2 , y = 1 + t, z = 2 – t, with parameters…
- If | a b | = 3 and | a b | = 4, then the angle between a and is
- The angle between the lines joining the points (1,0,-3),(2,-1,2) and (1,1,1),(3,2,0) respectively, is
- Vectors a ~b make an angle = 2 3 . If | a |=1,| ~b |=2 then ( a +3 b ) (3 a - b )^ 2 =
- If a line makes an angle of 4 with the positive directions of each of x - axis and y^ - axis, then the angle…
- A line passes through a point A with p.v. 3 i + j - k and is parallel to the vector 2 i - j +2 k . If P is a…
- For non-zero vectors a , b , c ,|( a b ) c |=| a || b || c | holds if and only if
- A line makes acute angles of α, β and γ with the co-ordinate axes such that cos α cos β = cos β cos = 2 9 and…
More Three Dimensional Geometry questions · Browse all practice questions