Question
If O is the origin and the coordinates of P are (2, 3,
β4), then find the equation of the plane passing through P and perpendicular to OP.Solution
The coordinates of the points, O and P, are (0, 0, 0) and (2, 3, β4) respectively. Therefore, the direction ratios of OP are (2 β 0) = 2, (3 β 0) = 3, and (β4 β 0) = β4 It is known that the equation of the plane passing through the point (x1, y1 z1) is Β a(x β x1) + b(y β y1) + c(z β z1) = 0 where, a, b, and c are the direction ratios of normal. Here, the direction ratios of normal are 2, 3, and β4 and the point P is (2, 3, β4). Thus, the equation of the required plane is 2(x β 2) + 3(y β 3) β 4(z + 4) = 0 => 2x + 3y β 4z β 29 = 0
- What will come in place of the question mark (?) in the following questions?
300β40%Β ofΒ 200=? - Simplify the following expression:
16 + [17 - (8 + 11) + 6 - 3] Γ· 0.2 Simplify the following expression.
(3-3 Γ 3 + 3 Γ· 3 + 3 Γ 5) Γ 2 of 5 + (2 + 2 Γ· 2 + 2 Γ 2 - 2)
2/9 of 5/8 of 3/25 of ? = 40
115 Γ· 23 + 12 Γ 6 = ? + 16 - 35
β? = 32% of 900 + 48% of 50
Β Β
5/13 Γ 104 + 1(2/9) Γ 198 = 133 + ?
Simplify the following expressions and choose the correct option.
40% of 360 + 25% of 248 - 30
(11/12) Γ (18/22) Γ (4/3) + 3 = ?2
[(15)³ × (8)²] ÷ (90 × 6) = ?²