# Solution - Equation of a Line in Space

Account
Register

Share

Books Shortlist

#### Question

A line passes through (2, −1, 3) and is perpendicular to the lines vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk) . Obtain its equation in vector and Cartesian from.

#### Solution

You need to to view the solution
Is there an error in this question or solution?

#### Similar questions VIEW ALL

The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.

view solution

If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.

view solution

The Cartestation equation of  line is (x-6)/2=(y+4)/7=(z-5)/3 find its vector equation.

view solution

Find the value of p, so that the lines l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5  are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.

view solution

Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines.

vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)

view solution

#### Reference Material

Solution for concept: Equation of a Line in Space. For the courses 12th CBSE (Arts), 12th CBSE (Commerce), 12th CBSE (Science)
S