Q91 / 155 · WBJEE 2026 · Mathematics
Let $\vec{a} = (x, y, z)$ be the vector with $|\vec{a}| = 2\sqrt{3}$, which makes equal angles with the vector $\vec{b} = (y, -2z, 3x)$ 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
- Correct answerA$(2, -2, -2)$
- B$(-2, -2, 2)$
- C$(-2, 2, -2)$
- D$(2, -2, 2)$
Solution
Given $\vec{a} = (x, y, z)$ and $|\vec{a}| = 2\sqrt{3}$, we have $x^2 + y^2 + z^2 = 12$.
Since $\vec{a}$ is perpendicular to $\vec{d} = (1, -1, 2)$, we have $\vec{a} \cdot \vec{d} = 0 \Rightarrow x - y + 2z = 0 \Rightarrow x - y = -2z$.
The angle between $\vec{a}$ and $\hat{j}$ is obtuse, so $\vec{a} \cdot \hat{j} < 0 \Rightarrow y < 0$.
The vector $\vec{a}$ makes equal angles with $\vec{b} = (y, -2z, 3x)$ and $\vec{c} = (2z, 3x, -y)$.
Notice that $|\vec{b}|^2 = y^2 + 4z^2 + 9x^2$ and $|\vec{c}|^2 = 4z^2 + 9x^2 + y^2$, so $|\vec{b}| = |\vec{c}|$.
Thus, $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$.
$x(y) + y(-2z) + z(3x) = x(2z) + y(3x) + z(-y)$
$xy - 2yz + 3zx = 2zx + 3xy - yz$
$zx - yz - 2xy = 0 \Rightarrow z(x - y) - 2xy = 0$
Substituting $x - y = -2z$ into the above equation:
$z(-2z) - 2xy = 0 \Rightarrow xy = -z^2$
Squaring $x - y = -2z$, we get:
$x^2 + y^2 - 2xy = 4z^2$
Substituting $xy = -z^2$ gives:
$x^2 + y^2 + 2z^2 = 4z^2 \Rightarrow x^2 + y^2 = 2z^2$
Using $x^2 + y^2 + z^2 = 12$, we get:
$2z^2 + z^2 = 12 \Rightarrow 3z^2 = 12 \Rightarrow z^2 = 4 \Rightarrow z = \pm 2$
For $z = -2$:
$xy = -4$ and $x - y = 4 \Rightarrow x = y + 4$
$(y + 4)y = -4 \Rightarrow y^2 + 4y + 4 = 0 \Rightarrow (y + 2)^2 = 0 \Rightarrow y = -2$
Since $y = -2 < 0$, this is valid. Then $x = 2$. The vector is $(2, -2, -2)$.
For $z = 2$:
$xy = -4$ and $x - y = -4 \Rightarrow x = y - 4$
$(y - 4)y = -4 \Rightarrow y^2 - 4y + 4 = 0 \Rightarrow (y - 2)^2 = 0 \Rightarrow y = 2$
Since $y = 2 > 0$, this is rejected.
Therefore, $\vec{a} = (2, -2, -2)$.
Answer: $(2, -2, -2)$
Since $\vec{a}$ is perpendicular to $\vec{d} = (1, -1, 2)$, we have $\vec{a} \cdot \vec{d} = 0 \Rightarrow x - y + 2z = 0 \Rightarrow x - y = -2z$.
The angle between $\vec{a}$ and $\hat{j}$ is obtuse, so $\vec{a} \cdot \hat{j} < 0 \Rightarrow y < 0$.
The vector $\vec{a}$ makes equal angles with $\vec{b} = (y, -2z, 3x)$ and $\vec{c} = (2z, 3x, -y)$.
Notice that $|\vec{b}|^2 = y^2 + 4z^2 + 9x^2$ and $|\vec{c}|^2 = 4z^2 + 9x^2 + y^2$, so $|\vec{b}| = |\vec{c}|$.
Thus, $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$.
$x(y) + y(-2z) + z(3x) = x(2z) + y(3x) + z(-y)$
$xy - 2yz + 3zx = 2zx + 3xy - yz$
$zx - yz - 2xy = 0 \Rightarrow z(x - y) - 2xy = 0$
Substituting $x - y = -2z$ into the above equation:
$z(-2z) - 2xy = 0 \Rightarrow xy = -z^2$
Squaring $x - y = -2z$, we get:
$x^2 + y^2 - 2xy = 4z^2$
Substituting $xy = -z^2$ gives:
$x^2 + y^2 + 2z^2 = 4z^2 \Rightarrow x^2 + y^2 = 2z^2$
Using $x^2 + y^2 + z^2 = 12$, we get:
$2z^2 + z^2 = 12 \Rightarrow 3z^2 = 12 \Rightarrow z^2 = 4 \Rightarrow z = \pm 2$
For $z = -2$:
$xy = -4$ and $x - y = 4 \Rightarrow x = y + 4$
$(y + 4)y = -4 \Rightarrow y^2 + 4y + 4 = 0 \Rightarrow (y + 2)^2 = 0 \Rightarrow y = -2$
Since $y = -2 < 0$, this is valid. Then $x = 2$. The vector is $(2, -2, -2)$.
For $z = 2$:
$xy = -4$ and $x - y = -4 \Rightarrow x = y - 4$
$(y - 4)y = -4 \Rightarrow y^2 - 4y + 4 = 0 \Rightarrow (y - 2)^2 = 0 \Rightarrow y = 2$
Since $y = 2 > 0$, this is rejected.
Therefore, $\vec{a} = (2, -2, -2)$.
Answer: $(2, -2, -2)$
Hence, the correct answer is option: 'A'.