To find the squared magnitude of the vector $\vec{CD}$, we first establish the parametric equations for lines $L_1$ and $L_2$, define the points of intersection C and D, and then use the condition that the vector $\vec{CD}$ is parallel to the direction vector of line $L_3$.
Line Equations Setup
Line $L_1$ passes through point $P_1 = (2, 6, 7)$ and is parallel to the vector $\vec{v_1} = -3\hat{i} + 2\hat{j} + 4\hat{k}$. Its parametric vector equation is:
$ \vec{r_1}(t) = (2\hat{i} + 6\hat{j} + 7\hat{k}) + t(-3\hat{i} + 2\hat{j} + 4\hat{k}) $
$ \vec{r_1}(t) = (2 - 3t)\hat{i} + (6 + 2t)\hat{j} + (7 + 4t)\hat{k} $
Line $L_2$ passes through point $P_2 = (4, 3, 5)$ and is parallel to the vector $\vec{v_2} = 2\hat{i} + \hat{j} + 3\hat{k}$. Its parametric vector equation is:
$ \vec{r_2}(s) = (4\hat{i} + 3\hat{j} + 5\hat{k}) + s(2\hat{i} + \hat{j} + 3\hat{k}) $
$ \vec{r_2}(s) = (4 + 2s)\hat{i} + (3 + s)\hat{j} + (5 + 3s)\hat{k} $
Vector $\vec{CD}$ Representation
Point C lies on $L_1$, so its coordinates are $C = (2 - 3t, 6 + 2t, 7 + 4t)$ for some parameter $t$.
Point D lies on $L_2$, so its coordinates are $D = (4 + 2s, 3 + s, 5 + 3s)$ for some parameter $s$.
The vector $\vec{CD}$ is calculated as $D - C$:
$ \vec{CD} = [(4 + 2s) - (2 - 3t)]\hat{i} + [(3 + s) - (6 + 2t)]\hat{j} + [(5 + 3s) - (7 + 4t)]\hat{k} $
$ \vec{CD} = (2 + 2s + 3t)\hat{i} + (-3 + s - 2t)\hat{j} + (-2 + 3s - 4t)\hat{k} $
Solving for Parameters $s$ and $t$
Line $L_3$ is parallel to the vector $\vec{v_3} = -3\hat{i} + 5\hat{j} + 16\hat{k}$. Since $L_3$ intersects $L_1$ at C and $L_2$ at D, the vector $\vec{CD}$ must be parallel to $\vec{v_3}$. This implies $\vec{CD} = k\vec{v_3}$ for some scalar $k$.
Equating the components:
Eliminate $k$ by combining equations. Multiply (1) by 5 and (2) by 3, then add:
$ 5(2 + 2s + 3t) + 3(-3 + s - 2t) = 5(-3k) + 3(5k) $
$ (10 + 10s + 15t) + (-9 + 3s - 6t) = -15k + 15k $
$ 1 + 13s + 9t = 0 \implies 13s + 9t = -1 $ (Equation A)
Multiply (1) by 16 and (3) by 3, then add:
$ 16(2 + 2s + 3t) + 3(-2 + 3s - 4t) = 16(-3k) + 3(16k) $
$ (32 + 32s + 48t) + (-6 + 9s - 12t) = -48k + 48k $
$ 26 + 41s + 36t = 0 \implies 41s + 36t = -26 $ (Equation B)
Solve the system formed by Equations A and B. Multiply Equation A by 4:
$ 4(13s + 9t) = 4(-1) \implies 52s + 36t = -4 $ (Equation A')
Subtract Equation B from Equation A':
$ (52s + 36t) - (41s + 36t) = -4 - (-26) $
$ 11s = 22 \implies s = 2 $
Substitute $s=2$ into Equation A:
$ 13(2) + 9t = -1 \implies 26 + 9t = -1 \implies 9t = -27 \implies t = -3 $
Find $k$ using Equation 1 with $s=2$ and $t=-3$:
$ 2 + 2(2) + 3(-3) = -3k \implies 2 + 4 - 9 = -3k \implies -3 = -3k \implies k = 1 $
- $ 2 + 2s + 3t = -3k $
- $ -3 + s - 2t = 5k $
- $ -2 + 3s - 4t = 16k $
Final Calculation of $| \vec{CD} |^2$
Since $k=1$, the vector $\vec{CD}$ is equal to $\vec{v_3}$:
$ \vec{CD} = 1 \times (-3\hat{i} + 5\hat{j} + 16\hat{k}) = -3\hat{i} + 5\hat{j} + 16\hat{k} $
The squared magnitude of $\vec{CD}$ is:
$ | \vec{CD} |^2 = (-3)^2 + (5)^2 + (16)^2 $
$ | \vec{CD} |^2 = 9 + 25 + 256 $
$ | \vec{CD} |^2 = 290 $
Similar Questions
- 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
- 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 :
- If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{7-z}{2}$ is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to
- Let $\vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha > \beta$, be the roots of the equation $\lambda^2 x^2 - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to
- Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \lambda \hat{j} + 2\hat{k}, \lambda \in \mathbb{Z}$ be two vectors. Let $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $yz$-plane. If $|\vec{c}| = \sqrt{53}$, then the maximum possible value of $(\vec{c} \cdot \vec{d})^2$ is equal to :
- Let L be the line $\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6}$ and let S be the set of all points $(a,b,c)$ on L, whose distance from the line $\frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0}$ along the line L is 7. Then $\sum_{(a,b,c)\in S} (a+b+c)$ 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 $(\alpha, \beta, \gamma)$ 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 :
- 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
- 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 _____.
Important Questions from Vectors and 3D Geometry
- 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
- 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 :
- If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{7-z}{2}$ is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to
- Let $\vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha > \beta$, be the roots of the equation $\lambda^2 x^2 - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to
- Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \lambda \hat{j} + 2\hat{k}, \lambda \in \mathbb{Z}$ be two vectors. Let $\vec{c} = \vec{a} \times \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $yz$-plane. If $|\vec{c}| = \sqrt{53}$, then the maximum possible value of $(\vec{c} \cdot \vec{d})^2$ is equal to :
