Question 28
Find the distance of point −2𝑖 ̂ + 3𝑗 ̂ − 4𝑘 ̂ from the line
𝑟 ̂ = 𝑖 ̂ + 2𝑗 ̂ − 𝑘 ̂ + 𝜆 (𝑖 ̂ + 3𝑗 ̂ − 9𝑘 ̂)
measured parallel to the plane: x – y + 2z – 3 = 0.
Let Point A = −2𝑖 ̂ + 3𝑗 ̂ − 4𝑘 ̂
∴ A = (–2, 3, –4)
Let point B the point on line such that AB is parallel to the plane
Equation of line is
𝑟 ̂ = 𝑖 ̂ + 2𝑗 ̂ − 𝑘 ̂ + 𝜆 (𝑖 ̂ + 3𝑗 ̂ − 9𝑘 ̂)
In Cartesian form,
(𝑥 − 1)/1=(𝑦 − 2)/3=(𝑧 + 1)/(−9)
Since point B lies on line
It’s coordinates will be
(𝑥 − 1)/1=(𝑦 − 2)/3=(𝑧 + 1)/(−9)=𝜆
So,
x = 𝜆+1
y = 3𝜆+2
z = −9𝜆−1
So, B = (𝜆+1, 3𝜆+2, −9𝜆−1)
Now,
Line AB is parallel to the plane
So, it will be perpendicular to plane’s normal
∴ Line AB is perpendicular to 𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂
Finding equation of line AB
Equation of a line passing through two points (x1, y1, z1) and (x1, y1, z1)
(𝑥 − 𝑥1)/(𝑥2 − 𝑥1) = (𝑦 − 𝑦1)/(𝑦2 − 𝑦1) = (𝑧 − 𝑧1)/(𝑧2 − 𝑧1)
We use (x1, y1, z1) = Point A = (–2, 3, –4)
& (x2, y2, z2) = B = (𝜆+1, 3𝜆+2, −9𝜆−1)
So, equation of AB is
(𝑥 − (−2))/((𝜆+1) − (−2)) = (𝑦 − 3)/((3𝜆+2) − (3)) = (𝑧 − (−4))/((−9𝜆−1) − (−4))
(𝑥 + 4)/(𝜆 + 3) = (𝑦 − 3)/(3𝜆 − 1) = (𝑧 + 4)/(−9𝜆 + 3)
From (1)
Line AB is perpendicular to 𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂
Direction cosines of AB = (𝜆+3, 3𝜆− 1, −9𝜆+3)
Direction cosines of 𝑖 ̂ – 𝑗 ̂ + 2𝑘 ̂ = (1, –1, 2)
Thus,
(𝜆+ 3) × 1 + (3𝜆− 1) × –1 + (−9𝜆+3) × 2 = 0
𝜆+ 3 – 3𝜆 + 1 −18𝜆+ 6 = 0
3 + 1 + 6 = 18𝜆+ 3𝜆 – 𝜆
10 = 20𝜆
10/20 = 𝜆
𝜆 = 1/2
So, point B is
B = (𝜆+1, 3𝜆+2, −9𝜆−1)
= (1/2 + 1, 3(1/2)+2, –9(1/2)−1)
= (3/2, 7/2,(−11)/2)
Thus,
Distance between A(–2, 3, –4) and B (3/2, 7/2,(−11)/2)
AB = √((3/2−(−2))^2+(7/2−3)^2+((−11)/2−(−4))^2 )
= √((3/2+2)^2+(7/2−3)^2+((−11)/2+4)^2 )
= √𝟓𝟗/𝟐 unit
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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