Question

Let the plane x+3y–2z+6=0 meet the co-ordinate axes at the points A,B,C. If the orthocentre of the triangle ABC is\( (α,β,\frac{6 }{7} ), \) then 98(α + β)² is equal to___.

Answer 1

Correct Answer: 288

The given points are:

\( A(-6, 0, 0), \quad B(0, -2, 0), \quad C(0, 0, 3) \)

Step 1: Vectors

\( \overrightarrow{AB} = 6\hat{i} - 2\hat{j} \)

\( \overrightarrow{BC} = 2\hat{j} + 3\hat{k} \)

\( \overrightarrow{AC} = 6\hat{i} + 3\hat{k} \)

Let \( H(\alpha, \beta, \frac{6}{7}) \) be the foot of the perpendicular from A to BC.

Step 2: Orthogonality of \( \overrightarrow{AH} \) and \( \overrightarrow{BC} \)

\( \overrightarrow{AH} \cdot \overrightarrow{BC} = 0 \)

Substituting:

\( (\alpha + 6, \beta, \frac{6}{7}) \cdot (0, 2, 3) = 0 \)

Simplify:

\( 2\beta + \frac{18}{7} = 0 \Rightarrow \beta = -\frac{9}{7} \)

Step 3: Orthogonality of \( \overrightarrow{CH} \) and \( \overrightarrow{AB} \)

\( \overrightarrow{CH} \cdot \overrightarrow{AB} = 0 \)

Substituting:

\( (\alpha, \beta, -\frac{15}{7}) \cdot (6, -2, 0) = 0 \)

Simplify:

\( 6\alpha - 2\beta = 0 \Rightarrow 6\alpha + \frac{18}{7} = 0 \Rightarrow \alpha = -\frac{3}{7} \)

Step 4: Calculating \( 98(\alpha + \beta)^2 \)

\( \alpha + \beta = -\frac{3}{7} - \frac{9}{7} = -\frac{12}{7} \)

Square the sum:

\( (\alpha + \beta)^2 = \left( -\frac{12}{7} \right)^2 = \frac{144}{49} \)

Multiply:

Quick Tip

\( 98(\alpha + \beta)^2 = 98 \cdot \frac{144}{49} = 2 \cdot 144 = 288 \)