JEE Mathematics 2023 Question with Solution

Given: The plane is

$$x + 3y - 2z + 6 = 0$$

and it meets the co-ordinate axes at points $A, B, C$. The orthocentre is $H(\alpha, \beta, \frac{6}{7})$.

Find: The value of $98(\alpha + \beta)^2$.

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 orthocentre.

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

$$\overrightarrow{AH} \cdot \overrightarrow{BC} = 0$$

Substituting,

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

So,

$$2\beta + \frac{18}{7} = 0$$

Hence,

$$\beta = -\frac{9}{7}$$

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

$$\overrightarrow{CH} \cdot \overrightarrow{AB} = 0$$

Substituting,

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

So,

$$6\alpha - 2\beta = 0$$

Using $\beta = -\frac{9}{7}$,

$$6\alpha + \frac{18}{7} = 0$$

Hence,

$$\alpha = -\frac{3}{7}$$

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

$$\alpha + \beta = -\frac{3}{7} - \frac{9}{7} = -\frac{12}{7}$$ $$(\alpha + \beta)^2 = \left(-\frac{12}{7}\right)^2 = \frac{144}{49}$$ $$98(\alpha + \beta)^2 = 98 \cdot \frac{144}{49} = 288$$

Therefore, the required value is $288$.

Given: $A(-6,0,0), B(0,-2,0), C(0,0,3)$ and orthocentre $H(\alpha,\beta,\frac{6}{7})$.

Find: $98(\alpha + \beta)^2$.

Since the orthocentre lies on the altitude from $A$, vector $\overrightarrow{AH}$ is perpendicular to $\overrightarrow{BC}$. Since it also lies on the altitude from $C$, vector $\overrightarrow{CH}$ is perpendicular to $\overrightarrow{AB}$.

So directly write:

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

which gives

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

Also,

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

which gives

$$6\alpha-2\beta=0 \Rightarrow \alpha=-\frac{3}{7}$$

Now,

$$\alpha+\beta=-\frac{12}{7}$$

Hence,

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

Therefore, the required value is $288$.