Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The reflection $P$ of point $Q$ across a line implies that the line acts as the perpendicular bisector of the segment $QP$, intersecting it at the midpoint $S$. Consequently, the area of $\triangle PQR$ is double the area of $\triangle QSR$, a relationship that significantly simplifies the calculation by allowing us to work with the smaller triangle $QSR$ formed by the projection $S$ and the given point $R$. To find $S$, we parameterize the line as $(t, 3+t, 1-t)$ and solve for the value of $t$ that makes the vector $\vec{QS}$ perpendicular to the line's direction vector $(1, 1, -1)$. Setting the dot product $(t-3, 6+t, -t) \cdot (1, 1, -1) = 0$ yields $t=-1$, identifying the projection point $S$ at $(-1, 2, 2)$.

With the coordinates of $S$ determined, we calculate the vectors $\vec{SQ} = (4, -5, -1)$ and $\vec{SR} = (3, 3, -3)$ to find the area of $\triangle QSR$ using the cross product. The cross product $\vec{SQ} \times \vec{SR}$ is given by the determinant of the matrix formed by the unit vectors and these components, resulting in:

$\vec{SQ} \times \vec{SR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -5 & -1 \\ 3 & 3 & -3 \end{vmatrix} = 18\hat{i} + 9\hat{j} + 27\hat{k}$

The magnitude of this cross product is $\sqrt{18^2 + 9^2 + 27^2} = 9\sqrt{4+1+9} = 9\sqrt{14}$. Because the area of $\triangle PQR$ is exactly the magnitude of this cross product, we have $\lambda = 9\sqrt{14}$. Squaring both sides gives $\lambda^2 = 81 \cdot 14$, and since the problem defines $\lambda^2 = 14K$, it follows that $K = 81$.

A point on a line is effectively defined using a single parameter, which allows us to express any point on the line as $(8\lambda-3, 2\lambda+4, 2\lambda-1)$ by setting $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}=\lambda$. To find the specific points at a distance of 6 units from $R(1,2,3)$, we apply the distance formula such that the square of the distance equals 36, resulting in the equation $(8\lambda-4)^{2+}(2\lambda+2)^{2+}(2\lambda-4)^2=36$. Simplifying this quadratic equation yields $72\lambda^{2-}72\lambda+36=36$, which reduces to $72\lambda(\lambda-1)=0$, identifying the parameters $\lambda=0$ and $\lambda=1$ for points $P(-3, 4, -1)$ and $Q(5, 6, 1)$, respectively. The centroid $(\alpha, \beta, \gamma)$ of triangle $PQR$ is then the average of the coordinates of $P$, $Q$, and $R(1,2,3)$, giving $\alpha=\frac{-3+5+1}{3}=1$, $\beta=\frac{4+6+2}{3}=4$, and $\gamma=\frac{-1+1+3}{3}=1$. The final value is calculated as $\alpha^{2+}\beta^{2+}\gamma^2=1^{2+}4^{2+}1^2=18$.

The distance of a point $P(x, y, z)$ from the x-axis is calculated as $\sqrt{y^2 + z^2}$, which is equivalent to $\sqrt{\gamma^2 - x^2}$ since the distance from the origin is $OP = \gamma$ and $x^2 + y^2 + z^2 = \gamma^2$. By considering the projection of $P$ onto the xy-plane as $OQ = \sqrt{x^2 + y^2}$, we identify that $\sin \phi = \frac{OQ}{\gamma}$ because $\phi$ is the angle between $OP$ and the z-axis, allowing us to express $OQ$ as $\gamma \sin \phi$. With $\theta$ defined as the angle between $OQ$ and the x-axis, the coordinate $x$ is equal to $OQ \cos \theta$, which substitutes as $\gamma \sin \phi \cos \theta$. Substituting this value for $x$ back into the distance formula yields $\sqrt{\gamma^2 - (\gamma \sin \phi \cos \theta)^2}$, which simplifies to $\gamma \sqrt{1 - \sin^2 \phi \cos^2 \theta}$.

The foot of the perpendicular from a point to a line is identified by the point $Q$ on the line where the vector connecting the given point and $Q$ is orthogonal to the direction of the line. The line passes through $(2, -5, 11)$ and $(-6, 7, -5)$, giving it a direction vector of $(-8, 12, -16)$, which simplifies to the proportional vector $(2, -3, 4)$. Using the point $(-6, 7, -5)$, the line is represented parametrically as $\frac{x+6}{2} = \frac{y-7}{-3} = \frac{z+5}{4} = \lambda$, allowing any point $Q$ on the line to be expressed as $(2\lambda - 6, 7 - 3\lambda, 4\lambda - 5)$.

Since $Q$ is the foot of the perpendicular from $R(1, 7, 6)$, the vector $\vec{RQ}$ formed by subtracting the coordinates of $R$ from $Q$ is $(2\lambda - 7, -3\lambda, 4\lambda - 11)$. This vector must be orthogonal to the direction vector $(2, -3, 4)$, meaning their dot product must be zero: $2(2\lambda - 7) - 3(-3\lambda) + 4(4\lambda - 11) = 0$. This expands to $4\lambda - 14 + 9\lambda + 16\lambda - 44 = 0$, which simplifies to $29\lambda - 58 = 0$, leading to $\lambda = 2$. Substituting $\lambda = 2$ into the parametric expression for $Q$ yields $(-2, 1, 3)$. The distance $PQ$ between $P(10, -2, -1)$ and $Q(-2, 1, 3)$ is found using the distance formula: $\sqrt{(-2-10)^2 + (1-(-2))^2 + (3-(-1))^2} = \sqrt{144 + 9 + 16} = 13$.

To find the equation of line $L$, we first parameterize the two given intersecting lines. Let any point $M$ on the line $\frac{x-2}{1} = \frac{y}{-1} = \frac{z-1}{1}$ be represented by coordinates $(\lambda+2, -\lambda, \lambda+1)$, and let any point $N$ on the line $\frac{x+1}{1/2} = \frac{y-1}{1/2} = \frac{z+1}{1}$ be represented by $(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1)$. Since line $L$ contains the segment $MN$ and is parallel to the vector $\langle 3, 1, 2 \rangle$, the direction vector $\vec{MN}$ must be proportional to this parallel direction. By setting the ratios of the components of $\vec{MN}$ proportional to $3, 1, 2$, we obtain the system $\frac{3+\lambda-\frac{\mu}{2}}{3} = \frac{-1-\lambda-\frac{\mu}{2}}{1} = \frac{2+\lambda-\mu}{2}$, which solves to $\lambda = -\frac{4}{3}$ and $\mu = -\frac{2}{3}$.

Substituting $\lambda = -\frac{4}{3}$ into the coordinate expression for $M$ yields the point $\left(\frac{2}{3}, \frac{4}{3}, -\frac{1}{3}\right)$. Using this point and the direction vector $\langle 3, 1, 2 \rangle$, the parametric equations for line $L$ are $x = \frac{2}{3} + 3k$, $y = \frac{4}{3} + k$, and $z = -\frac{1}{3} + 2k$. By evaluating these equations with $k = -\frac{1}{3}$, we determine the point $\left(\frac{2}{3} - 1, \frac{4}{3} - \frac{1}{3}, -\frac{1}{3} - \frac{2}{3}\right)$, which simplifies to $\left(-\frac{1}{3}, 1, -1\right)$.

The shortest distance between two skew lines, represented as $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$, is calculated using the scalar projection of the vector joining the two lines onto their common normal, expressed by the formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$. Extracting the relevant components from the given equations, the position vectors of points on the lines are $\vec{a}_1 = (3, -15, 9)$ and $\vec{a}_2 = (-1, 1, 9)$, while the direction vectors are $\vec{b}_1 = (2, -7, 5)$ and $\vec{b}_2 = (2, 1, -3)$.

Determining the displacement vector between the two points yields $\vec{a}_2 - \vec{a}_1 = (-4, 16, 0)$. Computing the cross product of the direction vectors provides the common normal vector

, which has a magnitude of $\sqrt{16^2 + 16^2 + 16^2} = 16\sqrt{3}$. Evaluating the dot product of the displacement vector and this normal vector results in $|(-4)(16) + (16)(16) + (0)(16)| = 192$. Dividing this value by the magnitude of the normal vector simplifies to $d = \frac{192}{16\sqrt{3}} = 4\sqrt{3}$.

The line of shortest distance between two skew lines is uniquely defined as the segment that intersects both lines and is perpendicular to both. The direction vector of this line is found by calculating the cross product of the direction vectors of the two given lines, $\vec{v}_1 = (-3, 4, 2)$ and $\vec{v}_2 = (-1, 2, 0)$. Their cross product, $\vec{v}_1 \times \vec{v}_2 = (4, 2, 2)$, simplifies to the proportional vector $(2, 1, 1)$, which dictates the orientation of the shortest distance line.

General points $P$ and $Q$ on the two respective lines can be expressed in terms of parameters as $P = (-3\lambda-2, 4\lambda+2, 2\lambda+5)$ and $Q = (-\mu-2, 2\mu-6, 1)$. Because the line connecting $P$ and $Q$ must be parallel to $(2, 1, 1)$, the components of the vector $\vec{PQ}$ are proportional to this direction vector. Setting up the equality of these ratios gives $\frac{3\lambda-\mu}{2} = \frac{2\mu-4\lambda-8}{1} = \frac{-2\lambda-4}{1}$. Solving these equations yields $\lambda = -1$ and $\mu = 1$.

With $\mu = 1$, the point $Q$ is $(-3, -4, 1)$. Using this point and the direction vector $(2, 1, 1)$, the equation of the shortest distance line is $\frac{x+3}{2} = \frac{y+4}{1} = \frac{z-1}{1}$. Since the point $(-1, \alpha, \beta)$ lies on this line, substituting $x = -1$ into the equation provides $\frac{-1+3}{2} = 1$, which implies $\frac{\alpha+4}{1} = 1$ and $\frac{\beta-1}{1} = 1$. Consequently, $\alpha = -3$ and $\beta = 2$, resulting in the final value $(\alpha - \beta)^2 = (-3 - 2)^2 = 25$.

Finding the image $(\alpha, \beta, \gamma)$ of point $(1, 0, 7)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ requires first defining a general point $M(\lambda, 1+2\lambda, 2+3\lambda)$ on the line. The vector connecting the initial point and this general point must be perpendicular to the direction of the given line, which implies that the dot product of $\vec{PM} = (\lambda-1, 1+2\lambda, 3\lambda-5)$ and the direction vector $(1, 2, 3)$ equals zero. This calculation leads to the equation $(\lambda-1) + 2(1+2\lambda) + 3(3\lambda-5) = 0$, which simplifies to $14\lambda = 14$ and yields $\lambda = 1$. The foot of the perpendicular is therefore $(1, 3, 5)$, and since this point serves as the midpoint between the initial point and its image $Q$, we calculate $Q = 2(1, 3, 5) - (1, 0, 7)$, resulting in $(\alpha, \beta, \gamma) = (1, 6, 3)$.

Once the coordinates of the image point are determined, the next step involves identifying the line passing through $(1, 6, 3)$ with the specified direction angles. Using the direction cosines $(l, m, n)$, where $m = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$ and $n = \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$, we satisfy the identity $l^{2+}m^{2+}n^2=1$. This leads to $l^2 + \frac{1}{4} + \frac{1}{2} = 1$, so $l^2 = \frac{1}{4}$, which implies $l = \frac{1}{2}$ given that the line forms an acute angle with the $x$-axis. The resulting line equation is represented by $\frac{x-1}{1/2} = \frac{y-6}{-1/2} = \frac{z-3}{-1/\sqrt{2}}$. Verifying the provided options shows that the point $(3, 4, 3-2\sqrt{2})$ aligns with this equation when the parameter is set to $4$.

The shortest distance between two skew lines can be determined by finding the projection of a vector connecting points on each line onto the vector perpendicular to both lines, which is their cross product. For the lines defined by $\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$ and $\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$, we define points on the lines as $P_1(4, -1, 0)$ and $P_2(\lambda, -1, 2)$ with direction vectors $\vec{v}_1 = \langle 1, 2, -3 \rangle$ and $\vec{v}_2 = \langle 2, 4, -5 \rangle$. The vector connecting these points is $\vec{P_1P_2} = \langle \lambda-4, 0, 2 \rangle$. Calculating the cross product $\vec{v}_1 \times \vec{v}_2$ gives the normal vector $\langle 2, -1, 0 \rangle$, which has a magnitude of $\sqrt{5}$.

The shortest distance is given by the formula $d = \frac{|\vec{P_1P_2} \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}$. Substituting our values yields $\frac{|\langle \lambda-4, 0, 2 \rangle \cdot \langle 2, -1, 0 \rangle|}{\sqrt{5}}$, which simplifies to $\frac{|2(\lambda-4)|}{\sqrt{5}}$. Setting this equal to the provided distance of $\frac{6}{\sqrt{5}}$ results in the equation $|2(\lambda-4)| = 6$, or simply $|\lambda-4| = 3$. This absolute value equation yields two possibilities, $\lambda - 4 = 3$ and $\lambda - 4 = -3$, resulting in $\lambda = 7$ and $\lambda = 1$. The sum of these values is 8.

Two lines are perpendicular if and only if the dot product of their direction vectors equals zero. Before performing this calculation, the given equations must be expressed in standard symmetric form, where the coefficients of the variables $x$, $y$, and $z$ are positive one. Rearranging the first line $\frac{2-x}{3} = \frac{3y-2}{4\lambda+1} = 4-z$ results in the form $\frac{x-2}{-3} = \frac{y-2/3}{(4\lambda+1)/3} = \frac{z-4}{-1}$, which identifies the first direction vector as $\langle -3, \frac{4\lambda+1}{3}, -1 \rangle$.

Applying the same standardization to the second line $\frac{x+3}{3\mu} = \frac{1-2y}{6} = \frac{5-z}{7}$ transforms it into $\frac{x+3}{3\mu} = \frac{y-1/2}{-3} = \frac{z-5}{-7}$, yielding the second direction vector $\langle 3\mu, -3, -7 \rangle$. Equating the dot product of these two vectors to zero produces the equation $(-3)(3\mu) + (\frac{4\lambda+1}{3})(-3) + (-1)(-7) = 0$. Simplifying this expression leads to $-9\mu - (4\lambda+1) + 7 = 0$, which reduces to $-9\mu - 4\lambda + 6 = 0$, confirming that $4\lambda + 9\mu = 6$.

The internal bisector of $\angle AOB$ divides the opposite side $AB$ into a ratio equal to the lengths of the adjacent sides $OA$ and $OB$. Since $|\vec{OA}| = 3$ and $|\vec{OB}| = 6$, the point $C$ divides $AB$ in a $1:2$ ratio. The section formula provides the position vector $\vec{OC} = \frac{1(2\hat{i}+4\hat{j}+4\hat{k}) + 2(2\hat{i}+2\hat{j}+\hat{k})}{1+2} = 2\hat{i}+\frac{8}{3}\hat{j}+2\hat{k}$. Squaring the components to determine the magnitude results in $\sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{\frac{136}{9}}$, which simplifies to $\frac{2}{3}\sqrt{34}$.

In a triangle with vertices $A(1,3,2)$, $B(-2,8,0)$, and $C(3,6,7)$, we first determine the side lengths to check for any special properties. Calculating the distance between these points shows that $AB = \sqrt{(-2-1)^2 + (8-3)^2 + (0-2)^2} = \sqrt{9+25+4} = \sqrt{38}$ and $AC = \sqrt{(3-1)^2 + (6-3)^2 + (7-2)^2} = \sqrt{4+9+25} = \sqrt{38}$. Since $AB = AC$, the triangle is isosceles, which means the angle bisector of $\angle BAC$ also functions as the median to side $BC$, placing the point $D$ at the midpoint of $BC$.

The coordinates of $D$ are found by averaging the coordinates of $B$ and $C$, yielding $D = \left( \frac{-2+3}{2}, \frac{8+6}{2}, \frac{0+7}{2} \right) = \left( \frac{1}{2}, 7, \frac{7}{2} \right)$. We define the vector $\vec{AD}$ as $D - A$, resulting in $\vec{AD} = \left( \frac{1}{2}-1 \right) \hat{i} + (7-3) \hat{j} + \left( \frac{7}{2}-2 \right) \hat{k} = -\frac{1}{2} \hat{i} + 4 \hat{j} + \frac{3}{2} \hat{k}$, and the vector $\vec{AC}$ as $C - A$, resulting in $\vec{AC} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$. The length of the projection of $\vec{AD}$ onto $\vec{AC}$ is given by the magnitude of their dot product divided by the length of $\vec{AC}$, which is $\left| \frac{\vec{AD} \cdot \vec{AC}}{|\vec{AC}|} \right|$. Computing this, the dot product is $\left( -\frac{1}{2} \times 2 \right) + (4 \times 3) + \left( \frac{3}{2} \times 5 \right) = -1 + 12 + 7.5 = \frac{37}{2}$, and the magnitude $|\vec{AC}|$ is $\sqrt{2^2 + 3^2 + 5^2} = \sqrt{38}$. Dividing these values gives the final result of $\frac{37}{2 \sqrt{38}}$.

The shortest distance between two skew lines is determined by the scalar projection of the vector connecting any two points on the lines onto the common perpendicular, which is the cross product of the direction vectors. Given the lines $\frac{x-\lambda}{3} = \frac{y-2}{-1} = \frac{z-1}{1}$ and $\frac{x+2}{-3} = \frac{y+5}{2} = \frac{z-4}{4}$, we identify the position vectors $\vec{a_1} = \lambda \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{a_2} = -2 \hat{i} - 5 \hat{j} + 4 \hat{k}$, alongside direction vectors $\vec{p} = 3 \hat{i} - \hat{j} + \hat{k}$ and $\vec{q} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$. The common perpendicular is found using the cross product $\vec{p} \times \vec{q} = -6 \hat{i} - 15 \hat{j} + 3 \hat{k}$, which has a magnitude of $\sqrt{(-6)^2 + (-15)^2 + 3^2} = 3\sqrt{30}$. Calculating the dot product of the difference vector $\vec{a_1} - \vec{a_2} = (\lambda+2) \hat{i} + 7 \hat{j} - 3 \hat{k}$ with the cross product vector yields $|-6\lambda - 126|$. Equating this geometric result to the provided shortest distance gives:

$\frac{|-6\lambda - 126|}{3\sqrt{30}} = \frac{44}{\sqrt{30}}$

Simplifying this expression leads to $|6\lambda + 126| = 132$. This absolute value equation yields two possible values for $\lambda$, where $6\lambda + 126 = 132$ results in $\lambda = 1$ and $6\lambda + 126 = -132$ results in $\lambda = -43$. Between these two possibilities, the largest absolute value $|\lambda|$ is 43.

The image of a point $A(8, 5, 7)$ in a given line is determined by finding the foot of the perpendicular $M$ from $A$ to that line, where $M$ serves as the midpoint between the original point and its image. Any point on the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5} = \lambda$ can be represented using the parameter $\lambda$ as $(2\lambda+1, 3\lambda-1, 5\lambda+2)$. The vector $\vec{AM}$ connecting the point to $M$ is $(2\lambda-7, 3\lambda-6, 5\lambda-5)$, and because this vector must be perpendicular to the line's direction vector $(2, 3, 5)$, their dot product must be zero. This relationship is written as:

$(2\lambda - 7)(2) + (3\lambda - 6)(3) + (5\lambda - 5)(5) = 0$

Solving this equation leads to $38\lambda = 57$, which simplifies to $\lambda = \frac{3}{2}$. By substituting this value back into the general coordinates for $M$, the foot of the perpendicular is found to be $\left(4, \frac{7}{2}, \frac{19}{2}\right)$. Given that $M$ is the midpoint of the segment connecting $A(8, 5, 7)$ and the image point $A'(\alpha, \beta, \gamma)$, we equate the coordinates: $\frac{\alpha+8}{2} = 4$, $\frac{\beta+5}{2} = \frac{7}{2}$, and $\frac{\gamma+7}{2} = \frac{19}{2}$. Calculating these results in $\alpha=0$, $\beta=2$, and $\gamma=12$, making the image point $(0, 2, 12)$. The sum of these coordinates is $0+2+12 = 14$.

The point of intersection $P$ is found by equating the coordinates of the two lines, representing them with parameters $\lambda$ and $\mu$. Setting the corresponding components equal gives the equations $\lambda+2 = 2\mu+3$ and $5\lambda+4 = 3\mu+2$. Solving this system leads to $\lambda = -1$ and $\mu = -1$, which reveals the intersection point $P(1, -1, 1)$.

The target line $4x=2y=z$ can be expressed in symmetric form as $\frac{x}{1} = \frac{y}{2} = \frac{z}{4} = k$. This implies a direction vector $\vec{v} = (1, 2, 4)$ and represents an arbitrary point $Q(k, 2k, 4k)$ on the line. The shortest distance occurs when the segment connecting $P$ to the line is perpendicular to the line itself. Representing this segment by the vector $\vec{PQ} = (k-1, 2k+1, 4k-1)$, the orthogonality condition $\vec{PQ} \cdot \vec{v} = 0$ requires that $1(k-1) + 2(2k+1) + 4(4k-1) = 0$.

Simplifying the dot product equation gives $k-1 + 4k+2 + 16k-4 = 0$, which results in $21k = 3$, or $k = \frac{1}{7}$. Using this value of $k$ allows for the calculation of the distance between point $P$ and the point $Q$ on the line:

$PQ = \sqrt{\left(1 - \frac{1}{7}\right)^2 + \left(-1 - \frac{2}{7}\right)^2 + \left(1 - \frac{4}{7}\right)^2} = \sqrt{\left(\frac{6}{7}\right)^2 + \left(-\frac{9}{7}\right)^2 + \left(\frac{3}{7}\right)^2} = \frac{\sqrt{36+81+9}}{7} = \frac{\sqrt{126}}{7} = \frac{3\sqrt{14}}{7}$

The centroid of a triangle serves as the geometric center that divides any median in a $2:1$ ratio, measured from the vertex to the midpoint of the opposite side. By applying the section formula to the segment connecting the given midpoint $M(2,1,2)$ and vertex $R(-1,4,2)$, where the centroid $G$ partitions the median $MR$ in a $1:2$ ratio closer to the midpoint, the coordinates of $G$ are calculated as $\frac{1(-1,4,2) + 2(2,1,2)}{3}$, resulting in $(1,2,2)$. Determining the point of intersection between the two lines requires parameterizing them and equating their coordinates, which yields the specific intersection point $A(2,-6,0)$. The Euclidean distance between $G(1,2,2)$ and $A(2,-6,0)$ is found using the distance formula $\sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2}$, which simplifies to $\sqrt{1 + 64 + 4}$ and results in $\sqrt{69}$.