To calculate \( c + 2d \), where \( C(c, d) \) represents the centroid of triangle \( \triangle PQR \) and the triangle's area is given as 35 square units, the following sequential procedure is employed:
\(Area = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\)
Given vertices are \( P(5, 4) \), \( Q(-2, 4) \), and \( R(a, b) \).
The area calculation yields:
\(35 = \frac{1}{2} \left| 5(4-b) + (-2)(b-4) + a(4-4) \right|\)
This equation simplifies to:
\(35 = \frac{1}{2} \left| 5(4-b) + 2(b-4) \right|\)
\(70 = \left| 20 - 5b + 2b - 8 \right|\)
\(70 = \left| 12 - 3b \right|\)
The equation \(70 = |12 - 3b|\) presents two possible scenarios:
\(c = \frac{5 - 2 + a}{3}\) and \(d = \frac{4 + 4 + b}{3}\)
The orthocenter \( O \) adheres to the relationship \( HO = 2 \times OG \), which establishes connections among \( O \), \( G \) (another representation of the centroid), and potentially \( H \) (the orthocenter).
The simplified centroid equations suggest potential values.
The final calculation yields values that indicate:
\( c + 2d = 3 \)
Consequently, the determined value is \( 3 \).
Foot of perpendicular from origin on a line passing through $(1, 1, 1)$ having direction ratios $\langle 2, 3, 4 \rangle$, is:
A line through $(1, 1, 1)$ and perpendicular to both $\hat{i} + 2\hat{j} + 2\hat{k}$ and $2\hat{i} + 2\hat{j} + \hat{k}$, let $(a, b, c)$ be foot of perpendicular from origin then $34 (a + b + c)$ is:
a times b is equal to