To find the possible values of \( \alpha \), we need to use the concept of reflection in a line in 3D geometry. The coordinates given are the image of the point through reflection in the line represented by \(\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}\).
First, identify the parametric equations of the line:
- The line in vector form is: \(\mathbf{r} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}\)
Given point: \((\alpha, 2\alpha, 1)\) and its image: \( (2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})\).
To obtain the reflection of a point in a line, determine the midpoint of the segment joining the point and its image and ensure it lies on the line.
Calculate the midpoint of the segment:
- Midpoint \((x_m, y_m, z_m)\) between a point \((x_1, y_1, z_1)\) and its image \((x_2, y_2, z_2)\) is: \((\frac{x_1 + x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})\).
Using this formula, the midpoint for our points is:
- \(x_m = \frac{\alpha + (2\alpha + 1)}{2} = \frac{3\alpha + 1}{2}\)
- \(y_m = \frac{2\alpha + (\alpha^2 - 3\alpha)}{2} = \frac{\alpha^2 - \alpha}{2}\)
- \(z_m = \frac{1 + \frac{\alpha - 1}{2}}{2} = \frac{2 + \alpha - 1}{4} = \frac{\alpha + 1}{4}\)
Since the midpoint lies on the line, it should satisfy the line's parametric equation:
- \(\frac{3\alpha + 1}{2} = 2 + 3t\)
- \(\frac{\alpha^2 - \alpha}{2} = 1 + 2t\)
- \(\frac{\alpha + 1}{4} = t\)
Solve the above three equations to determine the values of \( \alpha \):
- From the third equation: \( t = \frac{\alpha + 1}{4} \)
- Substitute \( t \) into the first equation:
\( \frac{3\alpha + 1}{2} = 2 + 3\left(\frac{\alpha + 1}{4}\right) \) simplifies to \( \alpha = 3 \).
(Checking this yields consistent solutions for all equations).
After checking these equations, only \( \alpha = 3 \) satisfies all conditions simultaneously. Thus, the possible value of \( \alpha \) is indeed Only 3.
Similar Questions
- If the image of the point P($a$, 2, $a$) in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is P, then $a+b$ is equal to _____.
- Let the area of the triangle formed by the lines $x+2=y-1=z$, $\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$ and $\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$ be $A$. Then $A^2$ is equal to
- The square of the distance of the point ($-2$, $-8$, 6) from the line $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{-1}$ along the line $\frac{x + 5}{1} = \frac{y + 5}{-1} = \frac{z}{2}$ is equal to:
- Let a vector $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}, \lambda > 0$, make an obtuse angle with the vector $\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$ and an angle $\theta, \frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive z-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta) - \{\gamma\}$, then $\alpha + \beta + \gamma$ is equal to ________.
Let $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a}$.
Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to :
- Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}|=1$, $|\vec{b}|=4$, $|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^\circ$, then $|\vec{a} \cdot \vec{c}|$ is equal to
- Let the point A be the foot of perpendicular drawn from the point $P(a, b, 0)$ on the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}$. If the midpoint of the line segment PA is $(0, \frac{3}{4}, \frac{-1}{4})$, then the value of $a^2 + b^2 + \alpha^2$ is equal to :
- Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $| \vec{CD} |^2$ is equal to :
- Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :
- Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point (5, 4, 2) on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is :
Important Questions from Vectors and 3D Geometry
- If the image of the point P($a$, 2, $a$) in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is P, then $a+b$ is equal to _____.
- Three vectors $\vec a,\vec b,\vec c$ are such that $|\vec a|=2,\ |\vec b|=2,\ |\vec c|=4$ and $\vec a+\vec b+\vec c=0$. Then the value of $4\vec a\cdot\vec b+3\vec b\cdot\vec c+3\vec c\cdot\vec a$ will be
- Let the area of the triangle formed by the lines $x+2=y-1=z$, $\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$ and $\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$ be $A$. Then $A^2$ is equal to
- A vector given by $\vec{P} = f(t)\hat{i} + g(t)\hat{j} + \hat{k}$ moves in such a way that it is always parallel to the vector $\vec{Q} = -f''(t)\hat{i} + f'(t)\hat{j} + \hat{k}$. The magnitude of $\vec{P}$ is
- The position vectors of two adjacent sides $\vec{OA}$ and $\vec{OB}$ of a rectangle $OACB$ are $\vec{a}$ and $\vec{b}$ respectively, where $O$ is the origin. If $16|\vec{a} \times \vec{b}| = 3(|\vec{a}| + |\vec{b}|)^2$ and $\theta$ be the acute angle between the diagonals $OC$ and $AB$, then the value of $\tan\left(\frac{\theta}{2}\right)$ is
