Example 13

Transcript

Example 13 Find the vector equation of the plane which is at a distance of 6﷮ ﷮29﷯﷯ from the origin and its normal vector from the origin is 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯. 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﷯) Now, distance from origin = d = 6﷮ ﷮29﷯﷯ 𝑛﷯ = 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯ Magnitude of 𝑛﷯ = ﷮22+ −3﷯﷮2﷯+ 4﷮2﷯﷯ 𝑛﷯﷯ = ﷮4+9+16﷯ = ﷮29﷯ Now, 𝑛﷯ = 1﷮ 𝑛﷯﷯﷯ ( 𝑛﷯) = 1﷮ ﷮29﷯﷯ (2 𝑖﷯−3 𝑗﷯+4 𝑘﷯) = 2﷮ ﷮29﷯﷯ 𝑖﷯ − 3﷮ ﷮29﷯﷯ 𝑗﷯ + 4﷮ ﷮29﷯﷯ 𝑘﷯ Vector equation is 𝑟﷯. 𝑛﷯ = d 𝑟﷯ . 2﷮ ﷮29﷯﷯ 𝑖﷯− 3﷮ ﷮29﷯﷯ 𝑗﷯+ 4﷮ ﷮29﷯﷯ 𝑘﷯﷯ = 6﷮ ﷮29﷯﷯ Therefore, vector equation of the plane is 𝒓﷯ . 𝟐﷮ ﷮𝟐𝟗﷯﷯ 𝒊﷯− 𝟑﷮ ﷮𝟐𝟗﷯﷯ 𝒋﷯+ 𝟒﷮ ﷮𝟐𝟗﷯﷯ 𝒌﷯﷯ = 𝟔﷮ ﷮𝟐𝟗﷯﷯ Cartesian equation Equation of a plane in Cartesian form which is at a distance ‘d’ from the origin and has a normal vector 𝑛﷯ = 𝑎 𝑖﷯ + b 𝑗﷯ + c 𝑘﷯ is lx + my + nz = d where l, m, n are direction cosines of 𝑛﷯ l = 𝑎﷮ ﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ + 𝑐﷮2﷯﷯﷯ , m = 𝑏﷮ ﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ + 𝑐﷮2﷯﷯﷯ , n = 𝑐﷮ ﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ + 𝑐﷮2﷯﷯﷯ Distance form origin = d = 6﷮ ﷮29﷯﷯ 𝑛﷯ = 2 𝑖﷯ − 3 𝑗﷯ + 4 𝑘﷯ Comparing with 𝑛﷯ = 𝑎 𝑖﷯ + 𝑏 𝑗﷯ + 𝑐 𝑘﷯, 𝑎 = 2, b = −3, c = 4 Also, ﷮𝑎2+𝑏2+𝑐2﷯ = ﷮22+(− 3)2+42﷯ = ﷮4+9+16﷯ = ﷮29﷯ So, direction cosines are l = 2﷮ ﷮29﷯﷯ , m = − 3﷮ ﷮29﷯﷯ , n = 4﷮ ﷮29﷯﷯ ∴ Cartesian equations of plane is lx + my + nz = d 2﷮ ﷮29﷯﷯ x + − 3﷮ ﷮29﷯﷯﷯y + 4﷮ ﷮29﷯﷯ z = 6﷮ ﷮29﷯﷯ 2x − 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x − 3y + 4z = 6