Direction Vector of Line L
Line L is perpendicular to two given lines. Their direction vectors are $\vec{d_1} = <4, 1, 1>$ and $\vec{d_2} = <1, 1, 0>$. The direction vector $\vec{d_L}$ of line L is found using the cross product of $\vec{d_1}$ and $\vec{d_2}$.
$\vec{d_L} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix}$ $= \mathbf{i}(1 \cdot 0 - 1 \cdot 1) - \mathbf{j}(4 \cdot 0 - 1 \cdot 1) + \mathbf{k}(4 \cdot 1 - 1 \cdot 1)$ $= \mathbf{i}(-1) - \mathbf{j}(-1) + \mathbf{k}(3) = <-1, 1, 3>$
Line L Equation and Point Q
Line L passes through $P(1, 1, 1)$ and has direction vector $\vec{d_L} = <-1, 1, 3>$. The parametric equation for L is $\vec{r} = <1, 1, 1> + t<-1, 1, 3>$, or: $x = 1 - t$ $y = 1 + t$ $z = 1 + 3t$
Point Q is where line L intersects the yz-plane ($x=0$). Set $x=0$: $1 - t = 0 \implies t = 1$. Calculate Q coordinates using $t=1$: $Q = (1-1, 1+1, 1+3(1)) = (0, 2, 4)$.
Line L' Equation and Point R
Line L' is parallel to L, so its direction vector is also $\vec{d_L} = <-1, 1, 3>$. Line L' passes through $S(1, 0, -1)$. The parametric equation for L' is $\vec{r} = <1, 0, -1> + u<-1, 1, 3>$, or: $x = 1 - u$ $y = u$ $z = -1 + 3u$
Point R is where line L' intersects the yz-plane ($x=0$). Set $x=0$: $1 - u = 0 \implies u = 1$. Calculate R coordinates using $u=1$: $R = (1-1, 1, -1+3(1)) = (0, 1, 2)$.
Area of Parallelogram PQRS
The vertices of the parallelogram are $P(1, 1, 1)$, $Q(0, 2, 4)$, $R(0, 1, 2)$, $S(1, 0, -1)$. We use the vectors representing two adjacent sides originating from P: $\vec{PQ}$ and $\vec{PS}$.
- $\vec{PQ} = Q - P = <0-1, 2-1, 4-1> = <-1, 1, 3>$
- $\vec{PS} = S - P = <1-1, 0-1, -1-1> = <0, -1, -2>$
The area of the parallelogram is the magnitude of the cross product $||\vec{PQ} \times \vec{PS}||$.
$\vec{PQ} \times \vec{PS} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 3 \\ 0 & -1 & -2 \end{vmatrix}$ $= \mathbf{i}(1(-2) - 3(-1)) - \mathbf{j}((-1)(-2) - 3(0)) + \mathbf{k}((-1)(-1) - 1(0))$ $= \mathbf{i}(-2 + 3) - \mathbf{j}(2 - 0) + \mathbf{k}(1 - 0)$ $= <1, -2, 1>$
The area is $||<1, -2, 1>|| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
The square of the area is $(\sqrt{6})^2 = 6$.
Similar Questions
- If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:
- Let $P(10, 2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are S and $S'$. If the length of its latus rectum is 8, then the square of the area of $\Delta PSS'$ is equal to :
- Let the locus of the mid-point of the chord through the origin O of the parabola $y^2 = 4x$ be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3:1, is :
- Let a circle of radius 4 pass through the origin O, the points $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab \neq 0$. Then the locus of the centroid of $\Delta OAB$ is a circle of radius
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1 + \theta_2)$ is:
- The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x + 1}{\alpha} = \frac{y - 2}{-1} = \frac{z - 4}{-\alpha}$ and $\frac{x}{\alpha} = \frac{y - 1}{2} = \frac{z - 1}{2\alpha}$ is $\sqrt{2}$, is
- Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$, then $(a^2 + b^2)$ is equal to
- Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
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- Let the image of parabola $x^2 = 4y$, in the line $x - y = 1$ be $(y + a)^2 = b(x - c)$, $a, b, c \in \mathbb{N}$. Then $a + b + c$ is equal to
Important Questions from Coordinate Geometry
- If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:
- Let $P(10, 2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are S and $S'$. If the length of its latus rectum is 8, then the square of the area of $\Delta PSS'$ is equal to :
- Let the locus of the mid-point of the chord through the origin O of the parabola $y^2 = 4x$ be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3:1, is :
- Let a circle of radius 4 pass through the origin O, the points $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab \neq 0$. Then the locus of the centroid of $\Delta OAB$ is a circle of radius
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1 + \theta_2)$ is:
