Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

Determining the intersection of two lines relies on expressing each coordinate $(x, y, z)$ on both lines in terms of independent parameters. For the first line passing through $(-1, 2, 1)$ with direction ratios $(2, 3, 4)$, the equation in symmetric form is $\frac{x+1}{2} = \frac{y-2}{3} = \frac{z-1}{4} = \lambda$, which provides a general point $(2\lambda-1, 3\lambda+2, 4\lambda+1)$. The second line, given by $\frac{x+2}{3} = \frac{y-3}{2} = \frac{z-4}{1} = \mu$, provides a corresponding general point $(3\mu-2, 2\mu+3, \mu+4)$.

Setting the components equal to find the intersection point $P$ leads to the system of equations $2\lambda - 1 = 3\mu - 2$, $3\lambda + 2 = 2\mu + 3$, and $4\lambda + 1 = \mu + 4$. Solving the first two equations simultaneously yields $\lambda = 1$ and $\mu = 1$, values that also satisfy the third equation, confirming the lines intersect at $P(1, 5, 5)$. The distance from this intersection point $P(1, 5, 5)$ to $Q(4, -5, 1)$ is found using the distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$, which results in $\sqrt{(1-4)^2 + (5 - (-5))^2 + (5-1)^2} = \sqrt{(-3)^2 + 10^2 + 4^2} = \sqrt{9 + 100 + 16} = \sqrt{125}$, simplifying to $5\sqrt{5}$.

For a line $L$ to pass through the point $(1,1,1)$ and intersect two given lines, $L_1$ and $L_2$, the points of intersection must form vectors with $(1,1,1)$ that are collinear with the direction vector of $L$. We identify any point on the first line, $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$, as $P_1(2\lambda+1, 3\lambda-1, 4\lambda+1)$ and any point on the second line, $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$, as $P_2(\mu+3, 2\mu+4, \mu)$. The vectors from the fixed point $(1,1,1)$ to $P_1$ and $P_2$ are given by $\vec{v_1} = \langle 2\lambda, 3\lambda-2, 4\lambda \rangle$ and $\vec{v_2} = \langle \mu+2, 2\mu+3, \mu-1 \rangle$, respectively.

Since these vectors must be proportional to define the same line $L$, we equate their components: $\frac{2\lambda}{\mu+2} = \frac{3\lambda-2}{2\mu+3} = \frac{4\lambda}{\mu-1}$. Solving this system of equations reveals that $\mu = -5$ and $\lambda = -6/5$. Substituting $\mu = -5$ into $\vec{v_2}$ yields the direction vector $\langle -3, -7, -6 \rangle$, which is parallel to $\langle 3, 7, 6 \rangle$. The equation of line $L$ passing through $(1,1,1)$ with this direction is $\frac{x-1}{3} = \frac{y-1}{7} = \frac{z-1}{6}$. Verifying the point $(7,15,13)$ in this equation confirms that $\frac{7-1}{3} = \frac{15-1}{7} = \frac{13-1}{6} = 2$, proving that this point lies on the line.

The line of shortest distance between two skew lines is their common perpendicular, which intersects each line at a specific point known as the foot of the perpendicular. To identify these points, consider an arbitrary point $P$ on the first line defined by $(2\lambda+1, 3\lambda+2, 4\lambda+3)$ and a point $Q$ on the second line defined by $(3\mu+2, 4\mu+4, 5\mu+5)$. The vector $\vec{PQ}$ must be parallel to the common perpendicular, which is the cross product of the direction vectors $(2, 3, 4)$ and $(3, 4, 5)$, resulting in $(-1, 2, -1)$.

Equating the direction ratios of $\vec{PQ}$-namely $(2\lambda-3\mu-1, 3\lambda-4\mu-2, 4\lambda-5\mu-2)$-to those of $(-1, 2, -1)$ yields the parameters $\lambda = 1/3$ and $\mu = -1/6$. These parameters locate the foot $P$ at $\left(\frac{5}{3}, 3, \frac{13}{3}\right)$, through which the line of shortest distance passes with a direction vector parallel to $\langle 1, -2, 1 \rangle$. The parametric equation for this line is $\frac{x-5/3}{1} = \frac{y-3}{-2} = \frac{z-13/3}{1} = k$, and setting the parameter $k=3$ results in the point $\left(3+\frac{5}{3}, -2(3)+3, 3+\frac{13}{3}\right)$, which simplifies to $\left(\frac{14}{3}, -3, \frac{22}{3}\right)$.

To determine the line $L$ that passes through $P(2, -1, 3)$ and is perpendicular to two other lines, we must first find its direction vector. This vector is obtained by calculating the cross product of the direction vectors of the two given lines, $(2, 1, -2)$ and $(1, 3, 4)$. This cross product yields $10\hat{i} - 10\hat{j} + 5\hat{k}$, which simplifies to the direction vector $(2, -2, 1)$. Given this direction and the point $P(2, -1, 3)$, the equation of the line $L$ can be expressed in terms of a parameter $\lambda$:

$\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda$

The line $L$ intersects the $yz$-plane at point $Q$, where the $x$-coordinate is $0$. By setting the expression for $x$, $x = 2\lambda + 2$, equal to $0$, we find that $2\lambda + 2 = 0$, which means $\lambda = -1$. Substituting this value back into the parameter equations for $y$ and $z$, we get $y = -2(-1) - 1 = 1$ and $z = -1 + 3 = 2$, identifying point $Q$ as $(0, 1, 2)$. The distance between $P(2, -1, 3)$ and $Q(0, 1, 2)$ is found using the distance formula, $\sqrt{(0-2)^2 + (1-(-1))^2 + (2-3)^2}$, which simplifies to $\sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

The shortest distance $d$ between two skew lines defined by points $\vec{a}_1, \vec{a}_2$ and direction vectors $\vec{b}_1, \vec{b}_2$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$. Given the first line passes through $\vec{a}_1 = -\hat{i}$ with direction $\vec{b}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}$ and the second line passes through $\vec{a}_2 = p\hat{i} + 2\hat{j} + \hat{k}$ with direction $\vec{b}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$, the cross product is $\vec{b}_1 \times \vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$, which has a magnitude of $\sqrt{6}$. Substituting these into the distance formula gives $d = \frac{|((p+1)\hat{i} + 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})|}{\sqrt{6}} = \frac{|p+1-4+1|}{\sqrt{6}} = \frac{|p-2|}{\sqrt{6}}$. Since $d = \frac{1}{\sqrt{6}}$, we have $|p-2| = 1$, resulting in $p=1$ and $p=3$. Because $a < b$, we set $a=1$ and $b=3$, creating the ellipse $\frac{x^2}{1} + \frac{y^2}{9} = 1$. The length of the latus rectum for an ellipse of the form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ is $\frac{2(\text{semi-minor})^2}{\text{semi-major}}$, which yields $\frac{2(1)^2}{3} = \frac{2}{3}$.

The shortest distance $d$ between two skew lines with point vectors $\vec{a}_1, \vec{a}_2$ and direction vectors $\vec{v}_1, \vec{v}_2$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}$. With $\vec{a}_1 = (3, \alpha, 3)$, $\vec{a}_2 = (-3, -7, \beta)$, $\vec{v}_1 = (3, -1, 1)$, and $\vec{v}_2 = (-3, 2, 4)$, the cross product is $\vec{v}_1 \times \vec{v}_2 = (-6, -15, 3)$, and its magnitude is $|\vec{v}_1 \times \vec{v}_2| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = 3\sqrt{30}$. Substituting these values into the distance formula $3\sqrt{30} = \frac{|(-6, -7-\alpha, \beta-3) \cdot (-6, -15, 3)|}{3\sqrt{30}}$ results in $|36 + 15(7+\alpha) + 3(\beta-3)| = 270$. Simplifying the absolute value expression yields $|15\alpha + 3\beta + 132| = 270$, which further simplifies to $|5\alpha + \beta + 44| = 90$. Solving for the positive value gives $5\alpha + \beta + 44 = 90$, resulting in $5\alpha + \beta = 46$.

Since the point $R(5, p, q)$ lies on the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$, the area of $\triangle PQR$ can be determined by identifying the perpendicular distance from $Q$ to the line and the segment distance along the line itself. The image $P$ of $Q$ in $L$ creates a triangle with a base $PQ$ that is perpendicular to $L$ at its midpoint $T$, which makes the length $RT$ the altitude of the triangle.

We express the coordinates of a general point on $L$ as $(2\lambda+1, 3\lambda-1, 4\lambda)$. For the specific point $R(5, p, q)$, setting the x-coordinate to $5$ gives $2\lambda+1=5$, which yields $\lambda=2$, so $R$ is $(5, 5, 8)$. To find the foot of the perpendicular $T$ from $Q(7,-2,5)$ onto $L$, we define $T$ as $(2\mu+1, 3\mu-1, 4\mu)$ such that the vector $\overrightarrow{QT} = (2\mu-6, 3\mu+1, 4\mu-5)$ is perpendicular to the direction vector $(2,3,4)$ of the line. The dot product $(2\mu-6)(2) + (3\mu+1)(3) + (4\mu-5)(4) = 0$ simplifies to $29\mu - 29 = 0$, giving $\mu=1$, which places $T$ at $(3, 2, 4)$.

The length $QT$ represents the perpendicular distance from $Q$ to the line, calculated as $\sqrt{(3-7)^2 + (2-(-2))^2 + (4-5)^2} = \sqrt{16+16+1} = \sqrt{33}$. Because $P$ is the image of $Q$ in $L$, $T$ is the midpoint of $PQ$, meaning the length of the base $PQ$ is $2QT = 2\sqrt{33}$. The height of the triangle from $R$ to the base $PQ$ is the distance $RT$ along the line, which is $\sqrt{(5-3)^2 + (5-2)^2 + (8-4)^2} = \sqrt{4+9+16} = \sqrt{29}$. Consequently, the area of $\triangle PQR$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2\sqrt{33}) \times \sqrt{29} = \sqrt{957}$, and the square of this area is 957.

To determine the coordinates of points $A$ and $B$, we use the geometric principle that if a single line passes through a fixed point $P(4, 1, 0)$ and intersects two other lines $L_1$ and $L_2$, then the points $P, A,$ and $B$ are necessarily collinear. This implies that the vectors $\vec{PA}$ and $\vec{PB}$ are proportional. Expressing any point on $L_1$ as $A(2\lambda+1, 3\lambda+2, 4\lambda+3)$ and any point on $L_2$ as $B(\mu+6, \mu, -\mu+4)$ in terms of parameters $\lambda$ and $\mu$, we define the vectors as $\vec{PA} = (2\lambda-3, 3\lambda+1, 4\lambda+3)$ and $\vec{PB} = (\mu+2, \mu-1, -\mu+4)$. Since $\vec{PA}$ and $\vec{PB}$ must be parallel, their components satisfy the ratio $\frac{2\lambda-3}{\mu+2} = \frac{3\lambda+1}{\mu-1} = \frac{4\lambda+3}{-\mu+4}$.

Cross-multiplying these ratios provides two equations: $\lambda\mu + 8\lambda + 4\mu = 1$ and $7\lambda\mu - 16\lambda + 4\mu = 7$. Solving this system, we find that the pair $\lambda = -1$ and $\mu = 3$ is a valid solution. Substituting $\lambda = -1$ into the parametric expression for $A$ gives $A(-1, -1, -1)$, and substituting $\mu = 3$ into the expression for $B$ gives $B(9, 3, 1)$.

With the coordinates of $A$ and $B$ identified, the required determinant is

. Expanding this along the first row, we compute $1((-1)(1) - (-1)(3)) - 0 + 1((-1)(3) - (-1)(9))$, which simplifies to $1(2) + 1(6)$, yielding a final value of $8$.

The distance from a point to a line is essentially the altitude of a right-angled triangle formed by the given point $P$, an arbitrary point $A$ on the line, and the projection point $Q$ on that line. By identifying the point $A(1, -2, -3)$ on the line and finding the vector $\vec{PA} = (-1, 8, -4)$ originating from the given point $P(2, -10, 1)$, we determine the length of the hypotenuse of this triangle as $|\vec{PA}| = \sqrt{(-1)^2 + 8^2 + (-4)^2} = 9$.

We then project $\vec{PA}$ onto the line's direction vector $\vec{v} = (2, -1, 2)$ to find the distance along the line, defined as $QA = \frac{|\vec{PA} \cdot \vec{v}|}{|\vec{v}|}$. Calculating the dot product $|(-1)(2) + (8)(-1) + (-4)(2)|$ yields $|-18|$, and dividing by the magnitude $\sqrt{2^{2+}(-1)^{2+}2^2} = 3$ gives a projection length of $6$. Applying the Pythagorean theorem, the square of the perpendicular distance $PQ$ is the difference between the squares of the hypotenuse and the base, $PQ^2 = 9^2 - 6^2 = 45$, resulting in $PQ = 3 \sqrt{5}$.

The shortest distance between two skew lines is determined by the projection of the vector connecting arbitrary points on the lines onto the vector perpendicular to both, which is the cross product of the lines' direction vectors. For line $L_1$ passing through $(1, 2, 3)$ parallel to the $z$-axis and line $L_2$ passing through $(\lambda, 5, 6)$ parallel to the $y$-axis, the direction vectors are $\hat{k}$ and $\hat{j}$, respectively, with their cross product being $\hat{k} \times \hat{j} = -\hat{i}$. A connecting vector between points on these lines is $(\lambda-1, 5-2, 6-3) = (\lambda-1, 3, 3)$, and the shortest distance is the magnitude of its projection onto $-\hat{i}$, which simplifies to $|\lambda-1|$. Setting this equal to 3 results in $\lambda = 4$ or $\lambda = -2$.

With $\lambda_2 < \lambda_1$, we identify $\lambda_2 = -2$ and $\lambda_1 = 4$, making the target point $(4, -2, 7)$. Line $L_1$ contains points $(1, 2, z)$, so identifying the distance from $(4, -2, 7)$ to $L_1$ requires the distance between $(4, -2, 7)$ and the point $(1, 2, 7)$ on the line, as the line maintains constant $x$ and $y$ values. The resulting distance calculation is:
$\sqrt{(4-1)^2 + (-2-2)^2 + (7-7)^2} = \sqrt{3^2 + (-4)^2} = 5$
Squaring this distance yields 25.

The direction vector of a line perpendicular to two given lines is found by taking the cross product of their individual direction vectors. For $L_1$ with direction $\langle 1, -1, 2 \rangle$ and $L_2$ with direction $\langle -1, 2, 1 \rangle$, their cross product yields the vector $\langle 5, 3, -1 \rangle$. Since the line $L_3$ intersects $L_1$ at the point $(\alpha, \beta, \gamma)$, this point must lie on $L_1$ and can be expressed in parametric form as $\alpha = p+1$, $\beta = -p+2$, and $\gamma = 2p+1$ for some scalar $p$.

Substituting these parametric forms of $\alpha$, $\beta$, and $\gamma$ into the expression $5\alpha - 11\beta - 8\gamma$ results in $5(p+1) - 11(-p+2) - 8(2p+1)$. Expanding these terms gives $5p + 5 + 11p - 22 - 16p - 8$. Combining the coefficients of $p$ results in $16p - 16p$, while the constant terms sum to $-25$. Consequently, the absolute value $\left|5\alpha - 11\beta - 8\gamma\right|$ evaluates to 25.

The area of triangle $PQR$ is determined by the formula $\frac{1}{2} \times \text{base} \times \text{height}$, where the length of segment $QR$ acts as the base and the perpendicular distance from vertex $P$ to the line serves as the altitude. Since points on the line follow the parametric form $x = 5\lambda - 3$, $y = 2\lambda + 1$, and $z = 3\lambda - 4$, we can represent any point $H$ on this line using the parameter $\lambda$.

The vector $\vec{PH}$ connecting $P(0, 2, 3)$ to an arbitrary point on the line must be orthogonal to the direction vector $(5, 2, 3)$ of the line. Setting the dot product equal to zero, $5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0$, results in $\lambda = 1$. Substituting this value provides the coordinates of the perpendicular foot $H(2, 3, -1)$, which yields a height $PH = \sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{21}$. Given the base $QR=5$ and height $\sqrt{21}$, the triangle area becomes $\frac{5\sqrt{21}}{2}$, so equating this to $\frac{m}{n}$ leads to the relation $2m - 5\sqrt{21}n = 0$.

The shortest distance between two skew lines in three-dimensional space is determined by projecting the vector connecting any two points on the lines onto the vector that is mutually perpendicular to both lines. We identify the displacement vector between the points $(7, 6, 2)$ and $(5, 3, 4)$ as $\vec{r} = (5-7)\hat{i} + (3-6)\hat{j} + (4-2)\hat{k} = -2\hat{i} - 3\hat{j} + 2\hat{k}$. We determine the common perpendicular direction using the cross product of the given direction vectors $\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$, which results in $\vec{n} = \vec{a} \times \vec{b} = 2\hat{i} + 17\hat{j} - 7\hat{k}$. Using the projection formula $d = \frac{|\vec{r} \cdot \vec{n}|}{|\vec{n}|}$, we calculate the dot product as $\vec{r} \cdot \vec{n} = -4 - 51 - 14 = -69$ and the magnitude $|\vec{n}| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{342}$. Simplifying the expression $\frac{|-69|}{\sqrt{342}}$ gives the final distance $\frac{23}{\sqrt{38}}$.

The shortest distance between two skew lines is determined by projecting the vector connecting any two points on the lines onto the common perpendicular, which is the cross product of the lines' direction vectors. Calculating the cross product for the given direction vectors $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$ yields the normal vector $\vec{n} = -\hat{i} + 2\hat{j} - \hat{k}$, which has a magnitude of $\sqrt{6}$. The distance formula becomes $\frac{|(\lambda-1)(-1) + 2(2) + 2(-1)|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$, which simplifies to $|3-\lambda| = 1$, identifying the values $\lambda = 2$ and $\lambda = 4$.

For the circle passing through the origin $(0,0)$ and the points $(4,2)$ and $(2,4)$, we use the circumradius formula $R = \frac{abc}{4\Delta}$. The triangle formed by these vertices has side lengths $2\sqrt{5}$, $2\sqrt{5}$, and $2\sqrt{2}$, and its area $\Delta$ is $6$. Substituting these values yields $R = \frac{(2\sqrt{5})(2\sqrt{5})(2\sqrt{2})}{4 \times 6} = \frac{40\sqrt{2}}{24}$, which simplifies to the radius $\frac{5\sqrt{2}}{3}$.

Finding the distance of a point from a line along a specific direction is equivalent to finding the intersection point of two lines. We construct a line passing through the given point $P(7, 10, 11)$ with the direction ratios $(2, 3, 6)$, which can be expressed parametrically as $\frac{x-7}{2} = \frac{y-10}{3} = \frac{z-11}{6} = \lambda$. Any point $Q$ on this line takes the coordinates $(2\lambda+7, 3\lambda+10, 6\lambda+11)$.

Because this point $Q$ must also lie on the target line $\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3}$, its coordinates must satisfy the second line's equation. Since the second line implies that the $y$-coordinate is fixed at $4$, we equate $3\lambda+10$ to $4$, which results in $\lambda = -2$. Substituting $\lambda = -2$ back into our parametric expressions gives the intersection point $Q(3, 4, -1)$. The distance between the original point $P(7, 10, 11)$ and the intersection point $Q(3, 4, -1)$ is calculated using the distance formula $\sqrt{(3-7)^2 + (4-10)^2 + (-1-11)^2}$, which simplifies to $\sqrt{16+36+144}$ and results in 14.

Finding the foot of the perpendicular $P$ from a given point to a line relies on the property that the vector formed by connecting the point to the foot must be orthogonal to the line's direction vector. By parameterizing an arbitrary point on line $L$ as $(\lambda+1, -\lambda-1, 2\lambda+2)$, we define the vector from $(1, 2, 2)$ to this general point as $(\lambda, -\lambda-3, 2\lambda)$. Setting the dot product of this vector and the line's direction vector $(1, -1, 2)$ to zero leads to the equation $1(\lambda) + (-1)(-\lambda-3) + 2(2\lambda) = 0$, which simplifies to $6\lambda + 3 = 0$. Solving for $\lambda$ gives $-1/2$, resulting in the coordinates of the foot $P$ as $(1/2, -1/2, 1)$.