Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).

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:

Point through which the required line passes: (−3, 5)
Line passes through points A(2, 5) and B(−3, 6)

Step 1: Find the slope of the given line AB

Slope of AB =

m_AB = (6 − 5) / (−3 − 2)

m_AB = 1 / (−5) = −1/5

Step 2: Find the slope of the required line

If two lines are perpendicular, then

m₁ · m₂ = −1

∴ Required slope =

m = 5

Step 3: Use point–slope form

Equation of line:

y − y₁ = m(x − x₁)

Substitute (x₁, y₁) = (−3, 5) and m = 5

y − 5 = 5(x + 3)

Step 4: Simplify

y − 5 = 5x + 15

y = 5x + 20

Final Answer:

The equation of the required line is:
y = 5x + 20

Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).

Solution and Explanation

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