Find the equation of the line that passes though \((2, 2\sqrt 3)\) and is inclined with the x-axis at an angle of 75°.

Question

If the components of \( \vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k} \) along and perpendicular to \( \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \) respectively are \( \frac{16}{11} (3\hat{i} + \hat{j} - \hat{k}) \) and \( \frac{1}{11} (-4\hat{i} - 5\hat{j} - 17\hat{k}) \), then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:

Answer 1

Given:

The line passes through the point (2, 2√3)
The line makes an angle of 75° with the positive x-axis

Step 1: Find the slope of the line

Slope of a line making an angle θ with the x-axis is:

m = tan θ

Here,

m = tan 75°

Using identity:

tan 75° = tan(45° + 30°)

tan 75° = (1 + 1/√3) / (1 − 1/√3) = 2 + √3

Step 2: Use point–slope form

Equation of a line:

y − y₁ = m(x − x₁)

Substitute (x₁, y₁) = (2, 2√3) and m = (2 + √3)

y − 2√3 = (2 + √3)(x − 2)

Final Answer:

The equation of the required line is:
y − 2√3 = (2 + √3)(x − 2)

Find the equation of the line that passes though \((2, 2\sqrt 3)\) and is inclined with the x-axis at an angle of 75°.

Solution and Explanation

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