The x-intercept of a line passing through the points $\left(-\frac{1}{2}, 1\right)$ and $(1, 2)$ is :

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

Step 1: Note the two points.
The line passes through $\left(-\frac{1}{2}, 1\right)$ and $(1, 2)$.

Step 2: Find the slope.
\[ m = \frac{2 - 1}{1 - \left(-\frac{1}{2}\right)} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]
Step 3: Write the line equation.
Use point $(1, 2)$ with slope $\frac{2}{3}$.
\[ y - 2 = \frac{2}{3}(x - 1) \]
Step 4: Set $y = 0$ for the x-intercept.
The x-intercept is where the line crosses the x-axis, so $y = 0$.
\[ 0 - 2 = \frac{2}{3}(x - 1) \]
Step 5: Solve for $x$.
\[ -2 = \frac{2}{3}(x - 1) \quad\Rightarrow\quad x - 1 = -3 \quad\Rightarrow\quad x = -2 \]
Step 6: Conclusion.
The line crosses the x-axis at $x = -2$. \[ \boxed{-2 \text{ (Option 2)}} \]

The x-intercept of a line passing through the points $\left(-\frac{1}{2}, 1\right)$ and $(1, 2)$ is :

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The Correct Option is B

Solution and Explanation

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