Condition for Two Lines to Be Perpendicular
Trending Questions
A line L is perpendicular to the line and the area of the triangle formed by line and coordinate axes is . The equation of the line is
From a point P(λ, λ, λ), perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and y = -x, z = -1. If P is such that ∠QPR is a right angle, then the possible value(s) of
λ is (are)
√2
1
-1
−√2
- Altitude through A is x−2y−9=0
- Altitude through B is 2x−y+12=0
- Altitude through C is x−y+1=0
- Orthocentre is (2, 4)
- x+2y−12=0
- x−2y−12=0
- x+2y+12=0
- x−2y+12=0
From a point P(λ, λ, λ), perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and y = -x, z = -1. If P is such that ∠QPR is a right angle, then the possible value(s) of
λ is (are)
√2
1
-1
−√2
- 6+√39
- 6−√39
- 12−√39
- 12+√39
- 16x+4y−5=0
- 16x−4y−5=0
- 16x+4y+5=0
- 16x−4y+5=0
- a=b
- 2a+b=0
- 3a−5b=0
- 5a−2b=0
- [−1√35, 5√35, 3√35]
- [−1√35, −5√35, 3√35]
- [1√35, 5√35, −3√35]
- [1√35, −7√35, −3√35]
- (m1n2−m2n1), (n1l2−l1n2), (l1m2−l2m1)
- (l1l2−m2m1), (m1m2−n1n2), (n1n2−l2l1)
- 1√l21+m21+n21, 1√l22+m22+n22, 1√3
- 1√3, 1√3, 1√3
- Altitude through A is x−2y−9=0
- Altitude through B is 2x−y+12=0
- Altitude through C is x−y+1=0
- Orthocentre is (2, 4)
- first
- second
- third
- fourth
A line l passing through the origin is perpendicular to the lines
l1:(3+t)^i+(−1+2t)^j+(4+2t)^k, −∞<t<∞l2:(3+2s)^i+(3+2s)^j+(2+s)^k, −∞<s<∞
Then, the coordinate(s) of the point(s) on l2 at a distance of √17 from the point of intersection of l and l1 is (are)
(73, 73, 53)
−1, −1, 0
(1, 1, 1)
(79, 79, 89)
- (m1n2−m2n1), (n1l2−l1n2), (l1m2−l2m1)
- (l1l2−m2m1), (m1m2−n1n2), (n1n2−l2l1)
- 1√l21+m21+n21, 1√l22+m22+n22, 1√3
- 1√3, 1√3, 1√3
- √23
- √211
- √25
- √27
- x+3y−10=0
- x+3y+10=0
- x−3y−10=0
- x−y+10=0
Then which of the following is correct ?
- a→q, s; b→r; c→p, q; d→p
- a→q; b→r; c→p, q; d→p, q
- a→s; b→r, s; c→p; d→p, q
- a→q, s; b→r, s; c→p; d→p
- x−21=y−15=z−1−1
- x−13=y−35=z−5−1
- x+22=y+1−1=z+11
- x−22=y−1−1=z−11
- 4x−3y+2=0
- 4x+3y+2=0
- 4x−3y−2=0
- 4x−3y+4=0
- −5
- 353
- 5
- −353
- 16x+4y−5=0
- 16x−4y−5=0
- 16x+4y+5=0
- 16x−4y+5=0
(-2, 5) is perpendicular to the line joining (6, 3) and (1, 1).
- True
- False
- ab′+bc′+1=0
- bb′+cc′+1=0
- aa′+c+c′=0
- cc′+a+a′=0
- x(1+tanθ)−y(1+tanθ)=py(1+tanθ)=x(1−tanθ)
- x(cosθ+sinθ)−y(cosθ+sinθ)=py(cosθ+sinθ)=x(cosθ+sinθ)
- x(cosθ−sinθ)+y(cosθ+sinθ)=py(cosθ−sinθ)=x(cosθ+sinθ)
- x(1+tan(π4+θ)−y(1+tan(π4+θ)=py(tan(π4+θ))=x(tan(π4−θ))
C(8, 12) and D(x, 24) is
- 4
- 1
- 6
- 8
- 12x+39y+155=0
- 12x−39y+155=0
- 12x+39y−155=0
- 12x−39y−155=0
A line l passing through the origin is perpendicular to the lines
l1:(3+t)^i+(−1+2t)^j+(4+2t)^k, −∞<t<∞l2:(3+2s)^i+(3+2s)^j+(2+s)^k, −∞<s<∞
Then, the coordinate(s) of the point(s) on l2 at a distance of √17 from the point of intersection of l and l1 is (are)
(73, 73, 53)
−1, −1, 0
(1, 1, 1)
(79, 79, 89)
- 122y−26x−1675=0
- 26x−122y−1675=0
- 26x+61y+1675=0
- 122y+26x+1675=0