Question

If the straight line passing through the points $(1,-1,2)$ and $(3,2,8)$ makes angle $\beta$ with the $y$-axis, then the value of $\cos \beta$ is equal to

• A. $\frac{1}{7}$
• B. $\frac{3}{7}$
• C. $\frac{2}{7}$
• D. $\frac{4}{7}$
• E. $\frac{5}{7}$

Hint:

The triplet (2, 3, 6) is very common in 3D problems because $2^2 + 3^2 + 6^2 = 49$, which is a perfect square ($7^2$). Memorizing this can speed up your calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The cosine of the angle a line makes with the y-axis is its \(y\)-direction cosine. To find this, we first need the direction ratios of the line connecting the two points.
Step 2: Key Formula or Approach:
The direction ratios \((a, b, c)\) of a line passing through \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are \(x_2 - x_1, y_2 - y_1, z_2 - z_1\).
The direction cosine with respect to the y-axis is \(\cos \beta = \frac{b}{\sqrt{a^2 + b^2 + c^2}}\).
Step 3: Detailed Explanation:
Let \((x_1, y_1, z_1) = (1, -1, 2)\) and \((x_2, y_2, z_2) = (3, 2, 8)\).
Calculate the direction ratios \((a, b, c)\):
\[ a = 3 - 1 = 2 \] \[ b = 2 - (-1) = 3 \] \[ c = 8 - 2 = 6 \] The direction ratios are \((2, 3, 6)\).
Now, compute the magnitude of this direction vector:
\[ \sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] The angle \(\beta\) is with the y-axis, so \(\cos \beta\) is the \(y\)-direction cosine:
\[ \cos \beta = \frac{b}{\sqrt{a^2 + b^2 + c^2}} = \frac{3}{7} \] Step 4: Final Answer:
The value of \(\cos \beta\) is \(\frac{3}{7}\).