Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

Any point $A$ on the line defined by $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-\alpha}{3}$ can be expressed in terms of a parameter $r$ as $(2r+1, r+2, 3r+\alpha)$. Since $\left(0, \frac{3}{4}, \frac{-1}{4}\right)$ is the midpoint of the segment $PA$ connecting $P(a, b, 0)$ to $A$, the coordinates of $A$ and $P$ relate to this midpoint through the standard midpoint formula. This allows for defining $a, b, \text{ and } \alpha$ as linear expressions involving $r$.

Because $A$ represents the foot of the perpendicular, the vector $\vec{PA}$ must be orthogonal to the direction vector of the line, which is $(2, 1, 3)$. Applying this perpendicularity condition enables the determination of the parameter $r=-\frac{1}{2}$. Solving the established equations with this value yields $a=0$, $b=0$, and $\alpha=1$. Consequently, the sum $a^{2+}b^{2+}\alpha^2$ evaluates to $1$.

The image $R$ of the point $P(0, -5, 0)$ in the line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$ is found by first locating the foot of the perpendicular $M$ on the line. Expressing any point on the line as $(2k+1, k, -2k-1)$, the vector $\vec{PM} = (2k+1, k+5, -2k-1)$ must be perpendicular to the line's direction vector $(2, 1, -2)$, leading to the equation $2(2k+1) + 1(k+5) - 2(-2k-1) = 0$. Solving this gives $k=-1$, which places $M$ at $(-1, -1, 1)$, and since $M$ is the midpoint of $PR$, $R$ is calculated as $2M-P$, resulting in $R(-2, 3, 2)$.

Similarly, the image $S$ of the point $Q(0, -1/2, 0)$ in the line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$ is determined by finding the foot of the perpendicular $L$. Setting the point on the line as $(-\lambda+1, 4\lambda-9, \lambda-1)$, the condition that the vector $\vec{QL} = (-\lambda+1, 4\lambda-17/2, \lambda-1)$ is perpendicular to the line's direction $(-1, 4, 1)$ results in $-1(-\lambda+1) + 4(4\lambda-17/2) + 1(\lambda-1) = 0$, which yields $\lambda=2$. This places $L$ at $(-1, -1, 1)$, and using the midpoint formula, the image $S$ is $2L-Q$, giving $S(-2, -3/2, 2)$.

The diagonals of the quadrilateral $PQRS$ are represented by vectors $\vec{PR} = (-2, 8, 2)$ and $\vec{QS} = (-2, -1, 2)$. Because the dot product of these diagonal vectors is $(-2)(-2) + (8)(-1) + (2)(2) = 0$, the diagonals are mutually perpendicular, confirming that $PQRS$ is a rhombus. The area of a rhombus is given by half the product of its diagonal lengths, which is $\frac{1}{2} |\vec{PR}| |\vec{QS}| = \frac{1}{2} (\sqrt{(-2)^2 + 8^2 + 2^2}) (\sqrt{(-2)^2 + (-1)^2 + 2^2}) = \frac{1}{2} (6\sqrt{2})(3) = 9\sqrt{2}$. Squaring this area yields $(9\sqrt{2})^2 = 162$.

The line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}=k$ provides points $Q$ with coordinates $(1+k, 2k, 1+k)$, making the vector $\vec{PQ}$ joining $P(1, 2, a)$ to $Q$ equal to $(k, 2k-2, k+1-a)$. For line $L_1$ with direction ratios $(3, 4, b)$, the proportionality condition $\frac{k}{3}=\frac{2k-2}{4}=\frac{k+1-a}{b}$ implies $k=3$ and $b=4-a$, resulting in a squared distance $PQ_1^2 = 3^{2+}4^{2+}b^2 = 25+(4-a)^2$. Similarly, for line $L_2$ with direction ratios $(1, 4, c)$, the proportionality condition $\frac{k}{1}=\frac{2k-2}{4}=\frac{k+1-a}{c}$ implies $k=-1$ and $c=a$, resulting in a squared distance $PQ_2^2 = (-1)^{2+}(-4)^{2+}(-a)^2 = 17+a^2$. Equating the squared distances $25+(4-a)^2 = 17+a^2$ simplifies to $25+16-8a+a^2 = 17+a^2$, which solves to $8a=24$, so $a=3$. Substituting $a=3$ gives $b=4-3=1$ and $c=3$, therefore $a+b+c = 3+1+3 = 7$.

The reflection of a point $P$ in a line is found by identifying the projection $M$ of $P$ onto that line, which acts as the midpoint between $P$ and its image $Q$, while ensuring the vector $\vec{PQ}$ is perpendicular to the line's direction vector. For point $P(a, 2, a)$ and line $L_1$ with direction $(2, 1, 1)$, the projection $R$ is found by setting the dot product of the vector from $P$ to a general point on $L_1$ and the direction vector to zero, which gives $k=\frac{2a+1}{3}$ for the parameter. This results in $R = (\frac{4a+2}{3}, \frac{1-a}{3}, \frac{2a+1}{3})$, and since $R$ is the midpoint of $PQ$, the image $Q$ is $2R - P = (\frac{5a+4}{3}, \frac{-2a-4}{3}, \frac{a+2}{3})$.

Since the reflection of $Q$ in the second line $L_2$ with direction $\vec{v_2}=(2, 1, -5)$ is $P$, the midpoint $M$ of $PQ$ must lie on $L_2$, and the vector $\vec{PQ} = (\frac{2a+4}{3}, \frac{-2a-10}{3}, \frac{2-2a}{3})$ must be perpendicular to $\vec{v_2}$. The condition $\vec{PQ} \cdot \vec{v_2} = 0$ simplifies to $12a-12=0$, yielding $a=1$. Substituting $a=1$ into $M$, we obtain $M(2, 0, 1)$, which must satisfy the equation of $L_2$ at $a=1$: $\frac{x-2b}{2} = \frac{y-1}{1} = \frac{z+2b}{-5}$. Setting the ratios equal at $M(2, 0, 1)$ gives $\frac{2-2b}{2} = -1$, which leads to $2-2b = -2$, so $b=2$. Therefore, $a+b = 1+2 = 3$.

Since the point $Q(1, \mu, 2)$ lies on the line defined by $\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$, its coordinates must satisfy the parametric form of this line. Substituting $x=1$ and $z=2$ into the equations gives $\frac{1-4}{1} = \frac{\mu-9}{2} = \frac{2-5}{1}$, which simplifies to $-3 = \frac{\mu-9}{2}$, resulting in $\mu=3$. Given that $Q$ is the foot of the perpendicular from $P(\lambda, 2, 3)$ to the line, the vector $\vec{PQ} = (1-\lambda)\hat{i} + (\mu-2)\hat{j} - \hat{k}$ must be orthogonal to the line's direction vector $\vec{d} = \hat{i} + 2\hat{j} + \hat{k}$. Calculating the dot product $\vec{PQ} \cdot \vec{d} = (1-\lambda)(1) + (2)(3-2) + (1)(-1) = 0$ simplifies to $1-\lambda+2-1=0$, which confirms that $\lambda=2$.

With $\lambda=2$ and $\mu=3$, the two given lines are $\vec{r_1} = (\hat{i}+2\hat{j}-4\hat{k}) + t_1(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r_2} = (2\hat{i}+3\hat{j}-5\hat{k}) + t_2(2\hat{i}+3\hat{j}+6\hat{k})$. Because these lines are parallel, the perpendicular distance between them is determined by the formula $\frac{|\vec{AB} \times \vec{v}|}{|\vec{v}|}$, where $\vec{AB}$ is the displacement vector between a point on the first line and a point on the second, and $\vec{v}$ is the shared direction vector. Using points $A(1, 2, -4)$ and $B(2, 3, -5)$, the vector connecting them is $\vec{AB} = \hat{i} + \hat{j} - \hat{k}$. The cross product

has a magnitude of $\sqrt{81+64+1} = \sqrt{146}$, and the magnitude of $\vec{v} = 2\hat{i}+3\hat{j}+6\hat{k}$ is $\sqrt{4+9+36} = 7$, leading to the final distance of $\frac{\sqrt{146}}{7}$.

The area of a triangle formed by vectors $\vec{AB}$ and $\vec{AC}$ is determined by the formula $\text{Area} = \frac{1}{2} |\vec{AB}| |\vec{AC}| \sin\theta$, where $\theta$ is the angle between the lines $L_1$ and $L_2$. To find the intersection point $A$, equate the coordinates of $L_1$ and $L_2$ such that $\lambda + 7 = 3\mu + 1$, $5 = 4\mu - 3$, and $-\lambda + 3 = 5\mu - 7$. Solving the equation $5 = 4\mu - 3$ gives $\mu = 2$, and substituting this into the $x$ coordinate yields $\lambda = 0$, confirming the intersection point is $A = (7, 5, 3)$.

The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = (1, 0, -1)$ and $\vec{v_2} = (3, 4, 5)$, respectively. The cosine of the angle $\theta$ between these lines is $\cos\theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|} = \frac{|(1)(3) + (0)(4) + (-1)(5)|}{\sqrt{1^{2+}0^{2+}(-1)^2}\sqrt{3^{2+}4^{2+}5^2}} = \frac{2}{\sqrt{2}\sqrt{50}} = \frac{2}{10} = \frac{1}{5}$. Applying the identity $\sin^2\theta = 1 - \cos^2\theta$, we obtain $\sin^2\theta = 1 - \left(\frac{1}{5}\right)^2 = \frac{24}{25}$.

The square of the area is calculated as $(\text{Area})^2 = \left(\frac{1}{2} |AB| |AC| \sin\theta\right)^2 = \frac{1}{4} (\sqrt{15})^2 (\sqrt{15})^2 \sin^2\theta = \frac{225}{4} \cdot \frac{24}{25}$. Simplifying this expression results in $9 \cdot 6 = 54$.

The area of a triangle defined by two vectors $\vec{PQ}$ and $\vec{PR}$ sharing a common vertex $P$ is given by half the magnitude of their cross product, area $= \frac{1}{2} | \vec{PQ} \times \vec{PR} |$. Because point $Q$ resides on a line passing through $P(-2, -1, 3)$ parallel to the vector $3\hat{i} + 2\hat{j} + 2\hat{k}$, its coordinates can be expressed in terms of a scalar $r$ as $(3r-2, 2r-1, 2r+3)$.

Applying the condition that the distance from $Q$ to $R(1, 3, 3)$ is 5, we solve the equation $(3r-2-1)^2 + (2r-1-3)^2 + (2r+3-3)^2 = 25$. Simplifying this leads to $(3r-3)^2 + (2r-4)^2 + (2r)^2 = 25$, which expands to $17r^2 - 34r + 25 = 25$, determining $r=2$ as the relevant solution for a distinct point $Q$. This places $Q$ at $(4, 3, 7)$, allowing us to determine the vectors $\vec{PQ} = 6\hat{i} + 4\hat{j} + 4\hat{k}$ and $\vec{PR} = 3\hat{i} + 4\hat{j} + 0\hat{k}$.

Computing the cross product $\vec{PQ} \times \vec{PR}$ results in $-16\hat{i} + 12\hat{j} + 12\hat{k}$. The square of the area is then calculated as $\left| \frac{1}{2}(-16\hat{i} + 12\hat{j} + 12\hat{k}) \right|^2$, which simplifies to the magnitude of the vector $-8\hat{i} + 6\hat{j} + 6\hat{k}$ squared. Evaluating this squared magnitude results in $(-8)^2 + 6^2 + 6^2 = 64 + 36 + 36$, which equals 136.

The vertices of the triangle are the points where each pair of lines intersects in three-dimensional space. Solving the system of linear equations formed by equating pairs of lines yields the coordinates $P(-2, 1, 0)$ from the first and second lines, $Q(0, 3, 2)$ from the first and third lines, and $R(3, 0, 1)$ from the second and third lines. These three points define the triangular region.

Representing the sides as vectors radiating from vertex $P$ gives $\vec{PQ} = 2\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{PR} = 5\hat{i} - \hat{j} + \hat{k}$. The area $A$ of a triangle formed by two vectors is given by $A = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$. Computing the cross product $\vec{PQ} \times \vec{PR}$ results in the vector $4\hat{i} + 8\hat{j} - 12\hat{k}$. The square of the magnitude of this resulting vector is $4^2 + 8^2 + (-12)^2 = 16 + 64 + 144 = 224$. Since $A^2 = \frac{1}{4} |\vec{PQ} \times \vec{PR}|^2$, the square of the area is $\frac{224}{4} = 56$.

The geometry of skew lines in three dimensions dictates that their shortest distance is found by projecting the displacement vector between any two points, one on each line, onto the common perpendicular vector. We identify the displacement vector between the points $(1, 2, 3)$ and $(0, 0, 5)$ as $\vec{a} - \vec{b} = (1, 2, -2)$. The direction vectors for the lines are $(2, 3, 4)$ and $(1, \alpha, 1)$, and their cross product provides the common perpendicular, which computes to $(3-4\alpha)\hat{i} + 2\hat{j} + (2\alpha-3)\hat{k}$.

Applying the distance formula requires the absolute value of the dot product of the displacement vector and the common perpendicular, normalized by the magnitude of that perpendicular:
$d = \left| \frac{(\vec{a} - \vec{b}) \cdot (\vec{p}_1 \times \vec{p}_2)}{|\vec{p}_1 \times \vec{p}_2|} \right| = \frac{5}{\sqrt{6}}$
Substituting the specific vectors into this formula leads to the expression $\frac{|13-8\alpha|}{\sqrt{20\alpha^{2-}36\alpha+22}} = \frac{5}{\sqrt{6}}$. Squaring both sides of this equation results in $6(13-8\alpha)^2 = 25(20\alpha^{2-}36\alpha+22)$, which simplifies to the quadratic equation $116\alpha^2 + 348\alpha - 464 = 0$. By applying Vieta's formulas to determine the sum of the roots for this quadratic, we calculate $-\frac{348}{116}$, which equals $-3$.

The area of any triangle is determined by half the product of its base and the perpendicular height. Given $AC = 6$, the task is to find the perpendicular distance from vertex $B(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$.

Any point $P$ on this line is written in terms of a parameter $\lambda$ as $(3\lambda+6, 2\lambda+7, -2\lambda+7)$. The vector $\vec{BP}$ from $B$ to $P$ is $(3\lambda+5, 2\lambda+5, -2\lambda+4)$. Because the height is the shortest distance, $\vec{BP}$ must be orthogonal to the line direction vector $(3, 2, -2)$, meaning their dot product is 0. This provides the equation $3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$, which simplifies to $17\lambda + 17 = 0$, identifying $\lambda = -1$.

Substituting $\lambda = -1$ into the expression for $\vec{BP}$ results in the vector $(2, 3, 6)$. The perpendicular height is the magnitude of this vector, $h = \sqrt{2^2 + 3^2 + 6^2} = 7$. With the base 6 and height 7, the area of the triangle is 21.

Since point $A$ lies in the $xy$-plane, its coordinates are $(x, y, 0)$. Because it is equidistant from the points $(0,3,2)$, $(2,0,3)$, and $(0,0,1)$, the squared distances to these points must be equal. Equating the squared distance from $A$ to $(0,0,1)$ with the squared distance from $A$ to $(0,3,2)$ simplifies to the condition:
$x^2 + y^2 + (0-1)^2 = x^2 + (y-3)^2 + (0-2)^2$
This calculation yields $y = 2$. By similarly equating the squared distances to $(0,3,2)$ and $(2,0,3)$, we obtain $6y = 4x$. Substituting $y = 2$ provides $x = 3$, identifying $A$ as $(3, 2, 0)$.

Given $B(1,4,-1)$ and $C(2,0,-2)$, the vectors $\vec{AB} = (-2, 2, -1)$ and $\vec{AC} = (-1, -2, -2)$ have a dot product of:
$\vec{AB} \cdot \vec{AC} = (-2)(-1) + (2)(-2) + (-1)(-2) = 2 - 4 + 2 = 0$
This result confirms that the triangle contains a right angle at $A$. The magnitudes are $AB = \sqrt{(-2)^2 + 2^2 + (-1)^2} = 3$ and $AC = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = 3$. Because $AB = AC$, the triangle is isosceles and right-angled, which validates statement S1. The area of the triangle is:
$\text{Area} = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$
As this result does not match the value provided in statement S2, only the first statement is true.

Two lines are perpendicular if the dot product of their direction vectors equals zero. The direction vector of the target line is $\langle -2, d, -4 \rangle$, while the direction vectors of the given lines are $\langle 1, a, b \rangle$ and $\langle -b, a, 5 \rangle$. Requiring the target line to be orthogonal to both yields two equations:
$-2 + ad - 4b = 0$
$2b + ad - 20 = 0$
Subtracting the first equation from the second eliminates $ad$ and results in $6b - 18 = 0$, which gives $b = 3$.

Because the line passes through the point $(0, -\frac{1}{2}, 0)$, these coordinates must satisfy the line equation $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$. Substituting $x=0$ and $y=-\frac{1}{2}$ into the first equality $\frac{0-1}{-2} = \frac{-\frac{1}{2}+4}{d}$ simplifies to $\frac{1}{2} = \frac{7}{2d}$, confirming $d = 7$. Substituting $x=0$ and $z=0$ into the relation $\frac{x-1}{-2} = \frac{z-c}{-4}$ gives $\frac{1}{2} = \frac{0-c}{-4}$, which simplifies to $\frac{1}{2} = \frac{c}{4}$, yielding $c = 2$. Finally, substituting $b=3$ and $d=7$ into the first orthogonality equation $-2 + 7a - 4(3) = 0$ results in $7a = 14$, so $a = 2$. The sum $a+b+c+d$ equals 14.

Any line in space is oriented by direction angles $\alpha, \beta, \gamma$ with respect to the coordinate axes, which inherently obey the identity $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given that the line makes equal angles with the $y$- and $z$-axes such that $\beta = \gamma$ and each is half the angle with the $x$-axis, substituting $\gamma = \beta$ and $\alpha = 2\beta$ into the identity yields the relationship:
$\cos^2 (2\beta) + 2\cos^2 \beta = 1$
Using the double-angle identity $\cos(2\beta) = 2\cos^2 \beta - 1$ transforms the equation into $(2\cos^2 \beta - 1)^2 + 2\cos^2 \beta = 1$, which further rearranges to:
$(2\cos^2 \beta - 1)^2 + (2\cos^2 \beta - 1) = 0$

Factoring the expression leads to $(2\cos^2 \beta - 1)(2\cos^2 \beta - 1 + 1) = 0$, which simplifies to $2\cos^2 \beta (2\cos^2 \beta - 1) = 0$. This product is zero if either $\cos^2 \beta = 0$ or $\cos^2 \beta = \frac{1}{2}$. These conditions correspond to $\beta = \frac{\pi}{2}$ and $\beta = \frac{\pi}{4}$, respectively. Adding these two possibilities, the sum of all potential values for the angle $\beta$ is $\frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.

To find the image of a point across a line, we first identify the foot of the perpendicular from the point to that line, as this foot serves as the midpoint between the original point and its reflected image. We start by expressing any point on the line passing through $A(4, 7, 1)$ and $B(3, 5, 3)$ using a parameter $k$. The direction vector of the line is $\vec{AB} = (3-4, 5-7, 3-1) = (-1, -2, 2)$, so any point on the line can be represented as $X = (4-k, 7-2k, 1+2k)$.

By defining a vector $\vec{PX}$ from $P(1, 0, 3)$ to an arbitrary point $X$ on the line, we apply the condition that the dot product of $\vec{PX}$ and the direction vector of the line must be zero for perpendicularity. This produces the equation $(3-k)(-1) + (7-2k)(-2) + (2k-2)(2) = 0$, which simplifies to $9k = 21$, or $k = \frac{7}{3}$. Substituting this parameter back into our expression for the line gives the coordinates of the foot of the perpendicular as $X = (\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$.

Because $X$ is the midpoint of $PQ$, we determine the coordinates of the image $Q(\alpha, \beta, \gamma)$ using the relation $Q = 2X - P$. Performing this operation yields $Q = (2(\frac{5}{3}) - 1, 2(\frac{7}{3}) - 0, 2(\frac{17}{3}) - 3)$, which simplifies to $(\frac{7}{3}, \frac{14}{3}, \frac{25}{3})$. Summing these coordinates gives $\alpha + \beta + \gamma = \frac{7}{3} + \frac{14}{3} + \frac{25}{3} = \frac{46}{3}$.

The distance from a point $P$ to a line $L$ in the direction of a vector $\vec{v}$ is determined by locating the unique point $Q$ on the line such that the vector $\vec{PQ}$ is parallel to $\vec{v}$, meaning $\vec{PQ} = \alpha \vec{v}$ for some scalar $\alpha$. Representing a general point $Q$ on the line $\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ via parameter $t$ as $(3t-1, 5t-3, 7t-5)$, the vector $\vec{PQ}$ from $P \left( \frac{15}{7}, \frac{32}{7}, 7 \right)$ is calculated as $\left( 3t - \frac{22}{7}, 5t - \frac{53}{7}, 7t - 12 \right)$.

Setting this displacement vector equal to the product of $\alpha$ and the given direction vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ allows for the formation of a system of equations. Solving these components consistently yields the scalar value $\alpha = -1$. Because the square of the distance corresponds to the squared magnitude of $\vec{PQ}$, the calculation becomes $|\alpha \vec{v}|^2 = \alpha^2 |\vec{v}|^2$, which simplifies to $(-1)^2 (1^2 + 4^2 + 7^2) = 1 + 16 + 49 = 66$.