1. Class 12
2. Important Question for exams Class 12

Transcript

Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) 2x + 3y + 4z – 12 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷﷯ is parallel to normal vector 𝒏﷯ to the plane. Given equation of plane is 2x + 3y + 4z − 12 = 0 2x + 3y + 4z = 12 Since, 𝑶𝑷﷯ and 𝒏﷯ are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 𝑎﷮1﷯﷮ 𝑎﷮2﷯﷯ = 𝑏﷮1﷯﷮ 𝑏﷮2﷯﷯ = 𝑐﷮1﷯﷮ 𝑐﷮2﷯﷯ = k 𝑥﷮1﷯﷮2﷯ = 𝑦﷮1﷯﷮3﷯ = 𝑧﷮1﷯﷮4﷯ = k x1 = 2k , y1 = 3k , z1 = 4k Also, point P(x1, y1, z1) lies in the plane. Putting P (2k, 3k, 4k) in 2x + 3y + 4z = 12, 2(2k) + 3(3k) + 4(4k) = 12 4k + 9k + 16k = 12 29k = 12 ∴ k = 12﷮29﷯ So, 𝑥﷮1﷯ = 2k = 2 × 12﷮29﷯ = 24﷮29﷯ 𝑦﷮1﷯ = 3k = 3 × 12﷮29﷯ = 36﷮29﷯ & 𝑧﷮1﷯ = 4k = 4 × 12﷮29﷯ = 48﷮29﷯ Therefore, coordinate of foot of perpendicular are 𝟐𝟒﷮𝟐𝟗﷯, 𝟑𝟔﷮𝟐𝟗﷯, 𝟒𝟖﷮𝟐𝟗﷯﷯ Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (b) 3y + 4z – 6 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷﷯ is parallel to normal vector 𝒏﷯ to the plane. Given equation of plane is 3y + 4z − 6 = 0 3y + 4z = 6 0x + 3y + 4z = 6 Since, 𝑶𝑷﷯ and 𝒏﷯ are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 𝑎﷮1﷯﷮ 𝑎﷮2﷯﷯ = 𝑏﷮1﷯﷮ 𝑏﷮2﷯﷯ = 𝑐﷮1﷯﷮ 𝑐﷮2﷯﷯ = k 𝑥﷮1﷯﷮0﷯ = 𝑦﷮1﷯﷮3﷯ = 𝑧﷮1﷯﷮4﷯ = k x1 = 0 , y1 = 3k , z1 = 4k Also, point P(x1, y1, z1) lies in the given plane. Putting P (0, 3k, 4k) in 0x + 3y + 4z = 6, 0 (k) + 3(3k) + 4(4k) = 6 25k = 6 k = 6﷮25﷯ So, 𝑥﷮1﷯ = 0 𝑦﷮1﷯ = 3k = 3 × 6 ﷮25﷯ = 18﷮25﷯ 𝑧﷮1﷯ = 4k = 4 × 6 ﷮25﷯ = 24﷮25﷯ Therefore, coordinates of foot of perpendicular are 𝟎, 𝟏𝟖﷮𝟐𝟓﷯, 𝟐𝟒﷮𝟐𝟓﷯﷯ Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (c) x + y + z = 1 Assume a point P (x1, y1, z1) on the plane. Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷﷯ is parallel to normal vector 𝒏﷯ to the plane. Given, equation of the plane is x + y + z = 1 1x + 1y + 1z = 1 Since, 𝑶𝑷﷯ and 𝒏﷯ are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 𝑎﷮1﷯﷮ 𝑎﷮2﷯﷯ = 𝑏﷮1﷯﷮ 𝑏﷮2﷯﷯ = 𝑐﷮1﷯﷮ 𝑐﷮2﷯﷯ = k 𝑥﷮1﷯﷮1﷯ = 𝑦﷮1﷯﷮1﷯ = 𝑧﷮1﷯﷮1﷯ = k x1 = y1 = z1 = k Also, point P(x1, y1, z1) lies in the given plane. Putting P (k, k, k) in x + y + z = 1, k + k + k = 1 3k = 1 ∴ k = 1﷮3﷯ So, 𝑥﷮1﷯= k = 1﷮3﷯ , 𝑦﷮1﷯ = k = 1﷮3﷯ , 𝑧﷮1﷯= k = 1﷮3﷯ Therefore, coordinate of foot of perpendicular are 𝟏﷮𝟑﷯, 𝟏﷮𝟑﷯, 𝟏﷮𝟑﷯﷯ Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (d) 5y + 8 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector 𝑶𝑷﷯ is parallel to normal vector 𝒏﷯ to the plane. Given, equation of plane 5y + 8 = 0 5y = − 8 − 5y = 8 0x − 5y + 0z = 8 Since, 𝑶𝑷﷯ and 𝒏﷯ are parallel their direction ratios are proportional. Finding direction ratios Since Direction ratios are proportional. 𝑎﷮1﷯﷮ 𝑎﷮2﷯﷯ = 𝑏﷮1﷯﷮ 𝑏﷮2﷯﷯ = 𝑐﷮1﷯﷮ 𝑐﷮2﷯﷯ 𝑥﷮1﷯﷮0﷯ = 𝑦﷮1﷯﷮ − 5﷯ = 𝑧﷮1﷯﷮0﷯ = k x1 = 0 , y1 = − 5k z1 = 0 Also, point P (x1, y1, z1) lies in the given plane. Putting (x1, y1, z1) in 0x − 5y + 0z = 8, 0x1 − 5y1 + 0z1 = 8, −5(−5k) = 8 25k = 8 ∴ k = 8﷮25﷯ So, x1 = 0, y1 = –5k = − 5 × 8﷮25﷯ = −8﷮5﷯ z1 = 0 ∴ Coordinate of foot of perpendicular is 0, −𝟖﷮𝟓﷯, 0﷯.

Class 12
Important Question for exams Class 12