


Get live Maths 1-on-1 Classs - Class 6 to 12
Examples
Example, 2 Important
Example, 3
Example, 4 Important
Example, 5 Important
Example, 6 Important
Example, 7
Example 8
Example, 9
Example 10 Important
Example 11
Example 12 Important
Example 13 Important Deleted for CBSE Board 2023 Exams
Example 14 Deleted for CBSE Board 2023 Exams
Example 15 Deleted for CBSE Board 2023 Exams
Example 16 Important Deleted for CBSE Board 2023 Exams You are here
Example 17 Deleted for CBSE Board 2023 Exams
Example 18 Deleted for CBSE Board 2023 Exams
Example 19 Important Deleted for CBSE Board 2023 Exams
Example 20 Important Deleted for CBSE Board 2023 Exams
Example 21 Important Deleted for CBSE Board 2023 Exams
Example 22 Deleted for CBSE Board 2023 Exams
Example 23 Important Deleted for CBSE Board 2023 Exams
Example 24 Deleted for CBSE Board 2023 Exams
Example, 25 Important Deleted for CBSE Board 2023 Exams
Example 26
Example 27 Important Deleted for CBSE Board 2023 Exams
Example 28 Important Deleted for CBSE Board 2023 Exams
Example 29 Important
Example 30 Important Deleted for CBSE Board 2023 Exams
Last updated at March 30, 2023 by Teachoo
Example 16 Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0. Let point P(x1, y1, z1) be foot of perpendicular from origin Since perpendicular to plane is parallel to normal vector Vector (𝑶𝑷) ⃗ is parallel to normal vector 𝒏 ⃗ Given equation of the plane is 2x − 3y + 4z − 6 = 0 2x − 3y + 4z = 6 So, Normal vector = 𝒏 ⃗ = 2𝒊 ̂ – 3𝒋 ̂ + 4𝒌 ̂ Since, (𝑶𝑷) ⃗ and 𝒏 ⃗ are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional 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 (𝑶𝑷) ⃗ = x1𝒊 ̂ + y1𝒋 ̂ + z1𝒌 ̂ Direction ratios = x1, y1, z1 ∴ a1 = x1 , b1 = y1, c1 = z1 𝒏 ⃗ = 2𝒊 ̂ – 3𝒋 ̂ + 4𝒌 ̂ Direction ratios = 2, −3, 4 ∴ a2 = 2 , b2 = –3, c2 = 4 Also, point P(x1, y1, z1) lies in the plane. Putting P (2k, − 3k, 4k) in equation of plane 2x − 3y + 4z = 6 2(2k) − 3(−3k) + 4(4k) = 6 4k + 9k + 16k = 6 29k = 6 ∴ k = 6/29 So, 𝑥_1 = 2k = 2 × 6/29 = 12/29 𝑦_1 = –3k = −3 × (6 )/29 = (−18)/29 𝑧_1 = 4k = 4 × 6/29 = 24/29 Therefore, coordinate of foot of perpendicular are (𝟏𝟐/𝟐𝟗, ( −𝟏𝟖)/𝟐𝟗,𝟐𝟒/𝟐𝟗)