Question

Let \[ \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}, \quad \text{where } a_1, a_2, a_3 \text{ and } |\vec{a}| \text{ are rational numbers.} \] If \(\vec{a}\) makes an angle of \(45^\circ\) with \[ \vec{b} = \sqrt{2} \hat{i} + 3 \sqrt{2} \hat{j} + 4 \hat{k}, \] then \(\vec{a}\) lies in which plane?

• A. XY-plane
• B. YZ-plane
• C. XZ-plane
• D. Along the bisector of the angle between \(\hat{k}\) and \(\vec{b}\)

Hint:

For vectors with rational magnitude and dot product, any irrational coefficient in one vector forces the corresponding component in the other vector to be zero.

Answer 1

The Correct Option is A

Step 1: Write the magnitude of b.
\[ |\vec b| = \sqrt{(\sqrt2)^2+(3\sqrt2)^2+4^2} = \sqrt{2+18+16} = \sqrt{36} = 6 \]

Step 2: Use the given angle to write the dot product.
\[ \vec a\cdot\vec b = |\vec a||\vec b|\cos45^\circ = 6|\vec a|\cdot\frac{1}{\sqrt2} = 3\sqrt2\,|\vec a| \]

Step 3: Expand the dot product in components and separate rational from irrational parts.
\[ \vec a\cdot\vec b = \sqrt2\,a_1 + 3\sqrt2\,a_2 + 4a_3 = \sqrt2(a_1+3a_2) + 4a_3 \] Setting this equal to \(3\sqrt2\,|\vec a|\) and rearranging:
\[ \sqrt2\big[(a_1+3a_2) - 3|\vec a|\big] + 4a_3 = 0 \]

Step 4: Separate the rational and irrational pieces.
Since \(a_1,a_2,a_3,|\vec a|\) are all rational, the bracketed term \((a_1+3a_2)-3|\vec a|\) is rational, while \(\sqrt2\) is irrational. For the whole sum to equal the rational number \(-4a_3/\sqrt2\)... in short, a rational multiple of \(\sqrt2\) plus a rational number can only be zero if both pieces vanish separately, so \(4a_3=0\), giving \(a_3=0\).

Step 5: Conclude the plane.
With \(a_3=0\), the vector \(\vec a\) has only \(x\) and \(y\) components.
\[ \boxed{\text{XY-plane}} \]