Question

Let \[ \vec{a}=(x,y,z) \] be the vector with \(|\vec{a}|=2\sqrt{3}\), which makes equal angles with the vectors \[ \vec{b}=(y,-2z,3x) \] and \[ \vec{c}=(2z,3x,-y), \] and is perpendicular to the vector \[ \vec{d}=(1,-1,2). \] If the angle between \(\vec{a}\) and the unit vector \(\hat{j}\) is obtuse, then \(\vec{a}\) is:

• A. $(2,-2,-2)$
• B. $(-2,-2,2)$
• C. $(-2,2,-2)$
• D. $(2,-2,2)$

Hint:

When evaluating a multi-constraint vector problem with explicit options, you can check the answer choices directly! Testing choice (B) $(-2,-2,2)$: its magnitude is $\sqrt{4+4+4}=2\sqrt{3}$, its dot product with $\vec{d}$ is $-2(1)-2(-1)+2(2) = -2+2+4 \neq 0$. Let's re-verify the component vector maps to confirm the exact bracket orientation choice.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves vector properties such as magnitude, dot product for perpendicularity, and the definition of angles between vectors. We are given the magnitude of \( \vec{a} \), conditions on its direction relative to other vectors, and a specific constraint regarding its direction relative to the y-axis.
Step 2: Key Formula or Approach:
1. Magnitude condition: \( x^2 + y^2 + z^2 = (2\sqrt{3})^2 = 12 \).
2. Perpendicularity with \( \vec{d} \): \( \vec{a} \cdot \vec{d} = 0 \implies x - y + 2z = 0 \).
3. Equal angles with \( \vec{b} \) and \( \vec{c} \): \( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} \).
4. Angle with \( \hat{j} \) is obtuse: \( \vec{a} \cdot \hat{j}<0 \implies y<0 \).
Step 3: Detailed Explanation:
First, check the magnitudes of \( \vec{b} \) and \( \vec{c} \):
\( |\vec{b}| = \sqrt{y^2 + (-2z)^2 + (3x)^2} = \sqrt{y^2 + 4z^2 + 9x^2} \)
\( |\vec{c}| = \sqrt{(2z)^2 + (3x)^2 + (-y)^2} = \sqrt{4z^2 + 9x^2 + y^2} \)
Since \( |\vec{b}| = |\vec{c}| \), the condition for equal angles simplifies to \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \).
\[ x(y) + y(-2z) + z(3x) = x(2z) + y(3x) + z(-y) \] \[ xy - 2yz + 3zx = 2zx + 3xy - yz \] \[ zx - 2xy - yz = 0 \quad \dots(1) \] From perpendicularity with \( \vec{d} \): \( x - y + 2z = 0 \implies y = x + 2z \).
Substitute \( y \) into (1):
\[ zx - 2x(x + 2z) - (x + 2z)z = 0 \] \[ zx - 2x^2 - 4xz - xz - 2z^2 = 0 \] \[ -2x^2 - 4xz - 2z^2 = 0 \implies x^2 + 2xz + z^2 = 0 \] \[ (x + z)^2 = 0 \implies z = -x \] Now substitute \( z = -x \) into \( y = x + 2z \):
\[ y = x + 2(-x) = -x \] So, \( x = -y = -z \). The vector is \( (x, -x, -x) \).
Use magnitude: \( x^2 + (-x)^2 + (-x)^2 = 12 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2 \).
If \( x = 2 \), then \( y = -2, z = -2 \). Vector is \( (2, -2, -2) \).
If \( x = -2 \), then \( y = 2, z = 2 \). Vector is \( (-2, 2, 2) \).
The condition \( y<0 \) (obtuse angle with \( \hat{j} \)) forces \( y = -2 \).
Thus, \( \vec{a} = (2, -2, -2) \).
Step 4: Final Answer:
The required vector is \( 2\hat{i} - 2\hat{j} - 2\hat{k} \).