Advertisements
Advertisements
प्रश्न
The equation of line passing through (3, –1, 2 ) and perpendicular to the lines
`vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and
`vecr = (2hati + hatj - 3hatk) + μ(hati - 2hatj + 2hatk)` is:
पर्याय
`(x + 3)/2 = (y + 1)/3 = (z - 2)/2`
`(x - 3)/3 = (y + 1)/2 = (z - 2)/2`
`(x - 3)/2 = (y + 1)/3 = (z - 2)/2`
`(x - 3)/2 = (y + 1)/2 = (z - 2)/3`
उत्तर
`bb((x - 3)/2 = (y + 1)/3 = (z - 2)/2)`
Explanation:
Let a, b and c be the direction ratios of the desired line which is also perpendicular to the given lines.
∴ 2a – 2b + c = 0
a – 2b + 2c = 0
`\implies a/|(-2, 1),(-2, 2)| = (-b)/|(2, 1),(1, 2)| = c/|(2, -2),(1, -2)|`
`\implies a/(-4 + 2) = (-b)/(4 - 1) = c/(-4 + 2)`
`\implies a/-2 = b/-3 = c/-2`
∴ Direction ratios are < –2, –3, –2 >
Hence equation of the desired line is
`(x - 3)/-2 = (y + 1)/-3 = (z - 2)/-2` or `(x - 3)/2 = (y + 1)/3 = (z - 2)/2`
