Example 14 - Class 12

Transcript

Example 14 Find the direction cosines of the unit vector perpendicular to the plane 𝑟﷯ .(6 𝑖﷯ − 3 𝑗﷯ – 2 𝑘﷯) + 1 = 0 passing through the origin. Vector equation of a plane at a distance ‘d’ from the origin and unit vector to normal from origin 𝑛﷯ is 𝒓﷯ . 𝒏﷯ = d Unit vector of 𝑛﷯ = 𝑛﷯ = 1﷮ 𝑛﷯﷯﷯( 𝑛﷯) Given, equation of plane is 𝑟﷯.(6 𝑖﷯ − 3 𝑗﷯ − 2 𝑘﷯) + 1 = 0 𝑟﷯.(6 𝑖﷯ − 3 𝑗﷯ − 2 𝑘﷯) = − 1 Multiplying with −1 on both sides, − 𝑟﷯.(6 𝑖﷯ − 3 𝑗﷯ − 2 𝑘﷯) = − 1 × −1 𝒓﷯. (–6 𝒊﷯ + 3 𝒋﷯ + 2 𝒌﷯) = 1 So; 𝑛﷯ = − 6 𝑖﷯ + 3 𝑗﷯ + 2 𝑘﷯ Magnitude of 𝑛﷯ = ﷮ − 6﷯2+32+22﷯ 𝑛﷯﷯ = ﷮36+9+4﷯ = ﷮49﷯ = 7 Now, 𝒏﷯ = 1﷮ 𝑛﷯﷯﷯ ( 𝑛﷯) = 1﷮7﷯ (− 6 𝑖﷯ + 3 𝑗﷯ + 2 𝑘﷯) = −𝟔﷮𝟕﷯ 𝒊﷯ + 𝟑﷮𝟕﷯ 𝒋﷯ + 𝟐﷮𝟕﷯ 𝒌﷯ ∴ Direction cosines of unit vector perpendicular to the given plane i.e. in are − 𝟔﷮𝟕﷯, 𝟑﷮𝟕﷯ , 𝟐﷮𝟕﷯ .