Question

The direction ratios of the normal to the plane passing through (1, 0, 0), (0, 1, 0) which makes an angle of measure \(45^{\circ}\) with the plane \(2x+3y = 7\) are....

• A. \(\sqrt{6}, 1, 1\)
• B. \(1, 1, \sqrt{6}\)
• C. \(\sqrt{13}, \sqrt{13}, \sqrt{6}\)
• D. \(\sqrt{13}, \sqrt{13}, 2\sqrt{6}\)

Hint:

The plane through (1,0,0) and (0,1,0) has normal (a, a, c); use the angle formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A plane \(ax + by + cz = d\) through (1, 0, 0) gives \(d = a\). Through (0, 1, 0) gives \(d = b\). So \(a = b\) and the normal is \((a, a, c)\).

Step 2: Apply the angle formula:
Normal of \(2x + 3y = 7\) is \((2, 3, 0)\), with magnitude \(\sqrt{13}\).
\[ \cos45^\circ = \frac{|2a + 3a|}{\sqrt{2a^2 + c^2}\sqrt{13}} = \frac{5a}{\sqrt{13}\sqrt{2a^2 + c^2}} \]

Step 3: Solve for c:
Squaring: \(\dfrac12 = \dfrac{25a^2}{13(2a^2 + c^2)}\), so \(13(2a^2 + c^2) = 50a^2\), giving \(13c^2 = 24a^2\), so \(c = a\sqrt{\dfrac{24}{13}}\).

Step 4: Direction ratios:
Scale by \(\sqrt{13}\): \((a, a, c)\to(\sqrt{13}, \sqrt{13}, \sqrt{24}) = (\sqrt{13}, \sqrt{13}, 2\sqrt6)\). Option (D).

Final Answer:
The normal has direction ratios root 13, root 13, 2 root 6. \[ \boxed{\text{(D) }\sqrt{13},\ \sqrt{13},\ 2\sqrt6} \]