Q.Find the vector equation of the line passing through the point (1,2,−4) and perpendicular to both the lines 3x−8=−16y+19=7z−10 and 3x−15=8y−29=−5z−5.
Concept understanding — Line Perpendicular To Two Lines
Line Perpendicular to Two Lines
In 3D geometry a very common task is this: two lines are given, and you must find the direction of a third line that is perpendicular to both of them. This shows up when finding a common perpendicular, the shortest distance between two lines, or the normal to a plane containing two directions.
The Core Idea
A line in space is fixed by two things: a point it passes through, and its direction vector. So "find a line perpendicular to two lines" really means "find a vector perpendicular to both of the two given direction vectors."
If the two given lines have direction vectors
b1=a1i^+b1j^+c1k^,b2=a2i^+b2j^+c2k^,
then a vector perpendicular to both is their cross product:
b1×b2=i^a1a2j^b1b2k^c1c2
Why the Cross Product?
The defining property of b1×b2 is that it is perpendicular to each factor:
(b1×b2)⋅b1=0,(b1×b2)⋅b2=0.
So it points in exactly the direction we need — along both perpendicularity conditions at once. This is why one cross product replaces solving a pair of dot-product equations by hand.
Note
The cross product only gives the direction of the perpendicular line. To pin down the actual line you still need a point it must pass through, given by the problem.
Using It
Suppose a line must be perpendicular to b1=i^+2j^+3k^ and b2=i^−j^+k^.
b1×b2=i^11j^2−1k^31=5i^+2j^−3k^.
So the required line has direction ratios ⟨5,2,−3⟩. Through a point A(x0,y0,z0) its equation is …
A line perpendicular to two given lines must be parallel to the cross product of their direction vectors, which fixes the required vector equation once combined with the given point. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2020Set 65/1/11 markMCQ
Q.The vector equation of the line passing through the point (−1,5,4) and perpendicular to the plane z=0 is
(A) r=−i^+5j^+4k^+λ(i^+j^)
(B) r=−i^+5j^+(4+λ)k^
(C) r=i^−5j^−4k^+λk^
(D) r=λk^
›Reveal solutionSolution
A line perpendicular to the plane z=0 must be parallel to the z-axis, so its direction vector is k^. The line passes through (−1,5,4), giving r=−i^+5j^+4k^+λk^. This matches option (B).
The plane z=0 is the xy-plane — a flat horizontal surface. Any line perpendicular to it must point straight up or down, i.e., parallel to the z-axis. That’s the core geometric insight.
A line’s vector equation is r=a+λd, where a is a point on the line and d is the direction vector. Here, the direction vector must be along k^ (or any scalar multiple of it). The given point is (−1,5,4), so a=−i^+5j^+4k^.
Now check each option:
Option (A):r=−i^+5j^+4k^+λ(i^+j^)
Direction is i^+j^, which lies in the xy-plane — parallel to z=0, not perpendicular. So this is wrong.
Option (B):r=−i^+5j^+(4+λ)k^
Rewrite as −i^+5j^+4k^+λk^. Direction is k^ — exactly what we need. This is correct.
Q.The direction cosines of two perpendicular lines are l₁, m₁, n₁ and l₂, m₂, n₂ respectively. Then the direction cosines of a line which is perpendicular to both these lines are:
(a) l₁ + l₂, m₁ + m₂, n₁ + n₂
(b) l₁ − l₂, m₁ − m₂, n₁ − n₂
(c) m₁n₂ − n₁m₂, n₁l₂ − l₁n₂, l₁m₂ − l₂m₁
(d) m₁n₁ − m₂n₂, l₁n₁ − l₂n₂, l₁m₂ − l₂m₁
›Reveal solutionSolution
A line perpendicular to two given lines is along their cross product.
If two lines have direction ratios (l₁,m₁,n₁) and (l₂,m₂,n₂), a vector perpendicular to both is given by the cross product of the two direction vectors: