Step 1: Finding the Foot of the Perpendicular
The line is given by the vector equation $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$.
A general point on the line has coordinates $(x, y, z)$, where:
- $x = -1 + 2\lambda$
- $y = 3 + 3\lambda$
- $z = 1 - \lambda$
Let the foot of the perpendicular from the point $P(5, 4, 2)$ be $F(\alpha, \beta, \gamma)$. This means $F$ is a point on the line, so $\alpha = -1 + 2\lambda$, $\beta = 3 + 3\lambda$, and $\gamma = 1 - \lambda$ for a specific value of $\lambda$.
The vector $\vec{PF}$ is calculated as $F - P$: $ \vec{PF} = (\alpha - 5)\hat{i} + (\beta - 4)\hat{j} + (\gamma - 2)\hat{k} $ Substituting the expressions for $\alpha, \beta, \gamma$: $ \vec{PF} = (-1 + 2\lambda - 5)\hat{i} + (3 + 3\lambda - 4)\hat{j} + (1 - \lambda - 2)\hat{k} $ $ \vec{PF} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-\lambda - 1)\hat{k} $
The direction vector of the line is $\vec{d} = 2\hat{i} + 3\hat{j} - \hat{k}$.
Since $\vec{PF}$ is perpendicular to the line's direction vector $\vec{d}$, their dot product must be zero: $\vec{PF} \cdot \vec{d} = 0$.
$ (2\lambda - 6)(2) + (3\lambda - 1)(3) + (-\lambda - 1)(-1) = 0 $ Expanding this equation: $ (4\lambda - 12) + (9\lambda - 3) + (\lambda + 1) = 0 $ Combining terms: $ 14\lambda - 14 = 0 $ Solving for $\lambda$: $ \lambda = 1 $
Now, we find the coordinates $(\alpha, \beta, \gamma)$ by substituting $\lambda = 1$ back into the general point equations:
- $ \alpha = -1 + 2(1) = 1 $
- $ \beta = 3 + 3(1) = 6 $
- $ \gamma = 1 - 1 = 0 $
Thus, the coordinates of the foot of the perpendicular are $(\alpha, \beta, \gamma) = (1, 6, 0)$.
Step 2: Calculating the Projection Length
We need to find the length of the projection of the vector $\vec{v} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ onto the vector $\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$.
Using the coordinates found, the vector $\vec{v}$ is $1\hat{i} + 6\hat{j} + 0\hat{k}$.
The formula for the length of the projection of vector $\vec{v}$ onto vector $\vec{w}$ is:
$ \text{Projection Length} = \frac{|\vec{v} \cdot \vec{w}|}{||\vec{w}||} $
Calculate the dot product $\vec{v} \cdot \vec{w}$:
$ \vec{v} \cdot \vec{w} = (1)(6) + (6)(2) + (0)(3) = 6 + 12 + 0 = 18 $
Calculate the magnitude of vector $\vec{w}$:
$ ||\vec{w}|| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 $
Now, compute the projection length:
$ \text{Projection Length} = \frac{|18|}{7} = \frac{18}{7} $
Similar Questions
- 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 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 $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a}$.
Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to :
- 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 _____.
- If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is 4, then the sum of all possible values of a 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 :
- 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:
- If $\left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right)$ is the image of $(\alpha, 2\alpha, 1)$ in the line $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}$, then the possible value(s) of $\alpha$ is (are)
- 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
Important Questions from Vectors and 3D Geometry
- 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 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 $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, $\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d} = \vec{c} \times \vec{a}$.
Then $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to :
- 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 _____.
