Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The plane $2x + my + nz = 4$ contains the point $(-2, \frac{7}{2}, \frac{3}{2})$, leading to the equation $2(-2) + m(\frac{7}{2}) + n(\frac{3}{2}) = 4$, which simplifies to $7m + 3n = 16$. Because the vector connecting $A(-1, 4, 3)$ to this point must be parallel to the plane's normal vector $(2, m, n)$, the components must satisfy the proportion $\frac{2}{-1} = \frac{m}{-1/2} = \frac{n}{-3/2}$. Solving this yields $m = 1$ and $n = 3$, establishing the plane equation as $2x + y + 3z = 4$.

To find the distance from $A$ to the plane measured parallel to the line with direction ratios $3, -1, -4$, we define the line passing through $A$ as $x = -1 + 3r, y = 4 - r, z = 3 - 4r$. Substituting these expressions into the plane equation results in $2(-1 + 3r) + (4 - r) + 3(3 - 4r) = 4$. Simplifying this linear equation gives $-7r + 11 = 4$, which implies $r = 1$. The required distance is the length of the segment along the line from $A$ to the intersection point, calculated as $\sqrt{(3(1))^2 + (-1)^2 + (-4(1))^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$.

The distance from a point to a plane corresponds to the length of the perpendicular segment dropped from that point onto the plane, acting as the altitude of the right-angled triangle $\triangle PQR$. For the point $P(1, 2, 3)$ and the plane $x+2y+z=14$, the length $PQ$ is calculated as
$PQ = \frac{|1(1) + 2(2) + 1(3) - 14|}{\sqrt{1^2 + 2^2 + 1^2}} = \frac{|-6|}{\sqrt{6}} = \sqrt{6}$

Because $PQ$ is normal to the plane, $\triangle PQR$ forms a right triangle with the right angle located at $Q$. Given the angle $\angle PRQ = 60^{\circ}$, the relationship between the sides is determined by the tangent function, where $\tan 60^{\circ} = \frac{PQ}{QR}$. Substituting the calculated length of $PQ$ results in $\sqrt{3} = \frac{\sqrt{6}}{QR}$, which simplifies to $QR = \sqrt{2}$. The area of $\triangle PQR$ is half the product of the legs $PQ$ and $QR$, yielding $\frac{1}{2} \times \sqrt{6} \times \sqrt{2} = \frac{\sqrt{12}}{2} = \sqrt{3}$.

The problem asks for the distance of a point from a plane along a prescribed path, which is solved by determining the intersection point $Q$ where the line passing through $P(3, 2, -1)$ with the given direction hits the plane. We rewrite the given line equation $\frac{2-x}{2} = \frac{y-3}{2} = \frac{z+1}{1}$ in standard form as $\frac{x-2}{-2} = \frac{y-3}{2} = \frac{z+1}{1}$, allowing us to represent any point on this line parametrically as $(3 - 2\lambda, 2 + 2\lambda, -1 + \lambda)$. For this point to lie on the plane $3x - y + 4z + 1 = 0$, substituting these parameters into the plane equation gives $3(3 - 2\lambda) - (2 + 2\lambda) + 4(-1 + \lambda) + 1 = 0$, which simplifies to $-4\lambda + 4 = 0$ and gives $\lambda = 1$. The intersection point $Q$ is therefore $(3 - 2(1), 2 + 2(1), -1 + 1)$, which simplifies to $(1, 4, 0)$. Finally, calculating the Euclidean distance between $P(3, 2, -1)$ and $Q(1, 4, 0)$ yields $\sqrt{(3 - 1)^2 + (2 - 4)^2 + (-1 - 0)^2} = \sqrt{4 + 4 + 1} = 3$.

The foot of the perpendicular from a point to a plane is the specific point on that plane where a line passing through the original point and perpendicular to the plane intersects it. Given the plane $-x+y+z=1$, the normal vector has components $-1$, $1$, and $1$. For any point $(x_0, y_0, z_0)$, the coordinates $(x', y', z')$ of its foot of perpendicular are found using the relation $\frac{x'-x_0}{-1} = \frac{y'-y_0}{1} = \frac{z'-z_0}{1} = \frac{-(-x_0+y_0+z_0-1)}{(-1)^{2+}1^{2+}1^2}$.

Substituting the point $P(1, 2, -1)$ into this relation gives a ratio of $1/3$. Equating the components of the line to this ratio, we find $x'-1 = -1/3$, $y'-2 = 1/3$, and $z'+1 = 1/3$, which identifies the foot of the perpendicular at $P'(2/3, 7/3, -2/3)$. Applying the same process for point $Q(2, -1, 3)$ also yields a ratio of $1/3$, resulting in $x''-2 = -1/3$, $y''+1 = 1/3$, and $z''-3 = 1/3$, which locates the foot of the perpendicular at $Q'(5/3, -2/3, 10/3)$.

The squared distance $d^2$ between $P'$ and $Q'$ is determined by the formula $(x''-x')^2 + (y''-y')^2 + (z''-z')^2$. Substituting these coordinates, we get $(5/3-2/3)^2 + (-2/3-7/3)^2 + (10/3-(-2/3))^2$. This simplifies to $1^2 + (-3)^2 + 4^2$, which equals $1 + 9 + 16$, totaling 26.

Finding the foot of a perpendicular from a point to a line relies on the geometric property that the vector connecting the given point and the foot $P$ must be orthogonal to the direction vector of the line. We start by expressing any point on the line $\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}$ as $(4t-2, 2t+1, 3t-1)$ in terms of a scalar parameter $t$. By calculating the vector from $(1, 2, 4)$ to this general point and taking its dot product with the line's direction vector $(4, 2, 3)$, we set the equation $4(4t-3) + 2(2t-1) + 3(3t-5) = 0$. Solving this linear equation yields $16t - 12 + 4t - 2 + 9t - 15 = 0$, or $29t = 29$, which confirms $t=1$. Substituting $t=1$ back into our parametric coordinates identifies the foot of the perpendicular $P$ as $(2, 3, 2)$.

With the point $P(2, 3, 2)$ determined, calculating its distance to the plane $3x+4y+12z+23=0$ involves applying the distance formula $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Substituting the coordinates of $P$ and the coefficients of the plane equation results in $\frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{3^2 + 4^2 + 12^2}}$. Evaluating the numerator gives $6 + 12 + 24 + 23 = 65$, while the denominator calculates to $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Dividing these values yields a final distance of 5.

A plane passing through a point $(x_0, y_0, z_0)$ and perpendicular to two given planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$ must have a normal vector collinear with the cross product of $\vec{n_1}$ and $\vec{n_2}$. This geometric property dictates that the equation of the plane satisfies:
$\begin{vmatrix} x-x_0 & y-y_0 & z-z_0 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}=0$
Using the point $(2,3,-5)$ and the normal vectors $(2, 1, -5)$ and $(3, 5, -7)$ from the given equations, the determinant is set up as:
$\begin{vmatrix} x-2 & y-3 & z+5 \\ 2 & 1 & -5 \\ 3 & 5 & -7 \end{vmatrix}=0$
Evaluating this determinant provides $(x-2)(-7 - (-25)) - (y-3)(-14 - (-15)) + (z+5)(10 - 3) = 0$, which simplifies to $18(x-2) - 1(y-3) + 7(z+5) = 0$. Expanding these terms results in $18x - 36 - y + 3 + 7z + 35 = 0$, or $18x - y + 7z + 2 = 0$. Rearranging to match the required form $ax+by+cz=d$ with $d > 0$ leads to $18x - y + 7z = -2$, and multiplying by $-1$ gives $-18x + y - 7z = 2$.

Comparing coefficients yields $a = -18$, $b = 1$, $c = -7$, and $d = 2$, which satisfies $\gcd(|-18|, 1, |-7|, 2) = 1$. The final value is found by substituting these into $a+7b+c+20d$, resulting in $-18 + 7(1) + (-7) + 20(2)$, which simplifies to $-18 + 7 - 7 + 40 = 22$.

Four points in space are coplanar if the scalar triple product of the vectors formed between them is zero. Setting point $A(2,3,9)$ as the origin for our vectors, we determine the vectors connecting $A$ to the other three points are $\vec{AB} = (3, -1, -8)$, $\vec{AC} = (-1, \lambda-3, -1)$, and $\vec{AD} = (\lambda-2, -1, -6)$.

The coplanarity condition is satisfied when the determinant of these vectors is zero:
$\begin{vmatrix} 3 & -1 & -8 \\ -1 & \lambda-3 & -1 \\ \lambda-2 & -1 & -6 \end{vmatrix} = 0$

Expanding the determinant results in the quadratic equation $8\lambda^2 - 57\lambda + 95 = 0$. By the properties of quadratic equations, the product of all possible values for $\lambda$ is given by the ratio of the constant term to the coefficient of the squared term, which yields $\frac{95}{8}$.

The intersection of two planes is a line whose direction vector must be simultaneously perpendicular to the normal vectors of both planes. By taking the cross product of the normal vectors of the two given planes, $ax + by = 3$ and $ax + by + cz = 0$, we obtain the direction vector of this line. Representing these normals as $\vec{n}_1 = (a, b, 0)$ and $\vec{n}_2 = (a, b, c)$, the cross product is calculated as $\vec{v} = \vec{n}_1 \times \vec{n}_2 = (bc, -ac, 0)$.

The angle $\theta$ between a line with direction vector $\vec{v}$ and a plane with normal vector $\vec{n}_3$ is determined by the relationship $\sin \theta = \frac{|\vec{v} \cdot \vec{n}_3|}{|\vec{v}| |\vec{n}_3|}$. Given the third plane $y - z + 2 = 0$, its normal vector is $\vec{n}_3 = (0, 1, -1)$. Substituting the known values, we have $\sin 30^\circ = \frac{1}{2} = \frac{|(bc, -ac, 0) \cdot (0, 1, -1)|}{\sqrt{(bc)^2 + (-ac)^2} \sqrt{0^2 + 1^2 + (-1)^2}}$.

Simplifying this equation, the numerator becomes $|-ac|$ and the denominator becomes $|c| \sqrt{a^2 + b^2} \sqrt{2}$. Since $a > 0$, the expression simplifies as shown:
$\frac{1}{2} = \frac{a}{\sqrt{2(a^2 + b^2)}}$
Squaring both sides results in $\frac{1}{4} = \frac{a^2}{2(a^2 + b^2)}$, which leads to $a^2 + b^2 = 2a^2$, confirming that $b^2 = a^2$ or $b = \pm a$. Substituting these values back into the direction vector $\vec{v} = (bc, -ac, 0)$, we find the possible direction ratios are proportional to $(ac, -ac, 0)$ or $(-ac, -ac, 0)$. Normalizing these vectors to unit length yields direction cosines of $\pm \frac{1}{\sqrt{2}}, \mp \frac{1}{\sqrt{2}}, 0$, which corresponds to the set $\frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}}, 0$.

Any plane passing through the intersection of two given planes, $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$, follows the linear combination $(5x + 8y + 13z - 29) + \lambda(8x - 7y + z - 20) = 0$. By rearranging this expression, the general equation becomes $(5+8\lambda)x + (8-7\lambda)y + (13+\lambda)z - (29+20\lambda) = 0$.

For plane $P_1$, which passes through $(2, 1, 3)$, substituting these coordinates into the general equation yields $(5+8\lambda)(2) + (8-7\lambda)(1) + (13+\lambda)(3) - (29+20\lambda) = 0$. Simplifying this results in $10 + 16\lambda + 8 - 7\lambda + 39 + 3\lambda - 29 - 20\lambda = 0$, which reduces to $-8\lambda + 28 = 0$, giving $\lambda = 7/2$. This value provides a normal vector $\vec{n_1} = \langle 1, -1/2, 1/2 \rangle$. Substituting $(0, 1, 2)$ for plane $P_2$ into the same general expression leads to $(5+8\lambda)(0) + (8-7\lambda)(1) + (13+\lambda)(2) - (29+20\lambda) = 0$, which simplifies to $8 - 7\lambda + 26 + 2\lambda - 29 - 20\lambda = 0$, resulting in $5 - 25\lambda = 0$, yielding $\lambda = 1/5$ and a normal vector $\vec{n_2} = \langle 1, 1, 2 \rangle$.

The acute angle $\theta$ between the two planes is determined by the dot product formula $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. The dot product $\vec{n_1} \cdot \vec{n_2} = (1)(1) + (-1/2)(1) + (1/2)(2) = 1 - 0.5 + 1 = 1.5$. The magnitudes are $|\vec{n_1}| = \sqrt{1^2 + (-1/2)^2 + (1/2)^2} = \sqrt{1.5}$ and $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$. Combining these results, $\cos \theta = \frac{1.5}{\sqrt{1.5} \sqrt{6}} = \frac{1.5}{\sqrt{9}} = 0.5$, which corresponds to an angle of $\pi/3$.

The set of points equidistant from two fixed locations in space forms a perpendicular bisector plane. By equating the squared distance from a variable point $(x, y, z)$ to $(-4, 2, 1)$ and to $(2, -2, 3)$, we have $(x+4)^2 + (y-2)^2 + (z-1)^2 = (x-2)^2 + (y+2)^2 + (z-3)^2$. Expanding these terms yields $x^2 + 8x + 16 + y^2 - 4y + 4 + z^2 - 2z + 1 = x^2 - 4x + 4 + y^2 + 4y + 4 + z^2 - 6z + 9$. Combining like terms leads to $12x - 8y + 4z + 4 = 0$, which simplifies to the equation $3x - 2y + z + 1 = 0$ for plane $P$.

The angle between two planes is defined by the angle between their normal vectors. The normal vector for plane $P$ is $\vec{n_1} = (3, -2, 1)$ and the normal vector for the plane $2x + y + 3z = 1$ is $\vec{n_2} = (2, 1, 3)$. The cosine of the angle $\theta$ is found using the dot product formula, where $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. Substituting the components, we get $\cos \theta = \frac{|(3)(2) + (-2)(1) + (1)(3)|}{\sqrt{3^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 3^2}}$, which calculates to $\frac{|6 - 2 + 3|}{\sqrt{14} \sqrt{14}}$. This results in $\frac{7}{14}$, or $1/2$, meaning the acute angle between the two planes is $\frac{\pi}{3}$.

The mirror image of a point across a plane lies on a line perpendicular to that plane, meaning the normal vector of the plane defines the direction of both the line segment connecting the point to its image and any line parallel to that segment. Given the plane $x+y+z=5$, the normal vector is $\langle 1,1,1 \rangle$, which acts as the direction vector for line $L$. By expressing a general point on line $L$ as $(\lambda+1, \lambda-1, \lambda-1)$ based on its passage through $(1,-1,-1)$ and direction $\langle 1,1,1 \rangle$, we substitute these coordinates into the plane equation to find $3\lambda - 1 = 5$, which results in $\lambda=2$ and the intersection point $R(3,1,1)$.

To locate the mirror image $Q$ of point $P(1,0,1)$, we identify a point along the normal from $P$ such that the line through $P$ and $Q$ is perpendicular to the plane. The equation for this line is $\frac{x-1}{1} = \frac{y}{1} = \frac{z-1}{1} = -2\frac{(1+0+1-5)}{1^{2+}1^{2+}1^2}$, which simplifies to $2$. Solving this equality for the coordinates yields $x=3$, $y=2$, and $z=3$, identifying $Q$ as $(3,2,3)$. Finally, the squared distance $QR^2$ is determined by the expression $(3-3)^2 + (2-1)^2 + (3-1)^2$, which evaluates to $5$.

The line of intersection between two planes is parallel to the cross product of their respective normal vectors. For the first plane, which is determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$, the normal vector is $\hat{k}$. For the second plane, determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$, the normal vector is obtained by the cross product $(\hat{i}-\hat{j}) \times (\hat{i}+\hat{k})$, which simplifies to $-\hat{i}-\hat{j}+\hat{k}$.

Calculating the cross product of these two normal vectors, $\hat{k} \times (-\hat{i}-\hat{j}+\hat{k})$, provides the direction vector $\overset{\rightarrow}{a} = \hat{i}-\hat{j}$ for the intersection line. To find the angle between this vector and $\overset{\rightarrow}{b} = \hat{i}-2\hat{j}+2\hat{k}$, we compute the dot product $\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b} = 1(1) + (-1)(-2) + 0(2) = 3$. Dividing this by the product of their magnitudes, $\left| \overset{\rightarrow}{a} \right| = \sqrt{2}$ and $\left| \overset{\rightarrow}{b} \right| = 3$, gives $\cos \theta = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$. This cosine value corresponds to an acute angle of $\frac{\pi}{4}$, meaning the required obtuse angle is $\pi - \frac{\pi}{4}$, which equals $\frac{3\pi}{4}$.

The shortest distance between two skew lines in space is the length of the common perpendicular segment connecting them, which is calculated as the magnitude of the projection of the vector joining any two points on the lines onto their common normal vector. By inspecting the line equations, we identify the position vectors for points on the lines, $\vec{a}_1 = 3\hat{i} + 2\hat{j} + 1\hat{k}$ and $\vec{a}_2 = -3\hat{i} + 6\hat{j} + 5\hat{k}$, alongside their direction vectors $\vec{v}_1 = 2\hat{i} + 3\hat{j} - 1\hat{k}$ and $\vec{v}_2 = 2\hat{i} + 1\hat{j} + 3\hat{k}$.

The vector connecting the two lines is found by taking the difference $\vec{a}_2 - \vec{a}_1 = -6\hat{i} + 4\hat{j} + 4\hat{k}$, while the common normal vector is determined by the cross product of the direction vectors:
$\vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix} = 10\hat{i} - 8\hat{j} - 4\hat{k}$
To determine the shortest distance, we take the absolute value of the dot product of the connecting vector and the common normal, then divide by the magnitude of the normal vector:
$d = \left| \frac{(-6\hat{i} + 4\hat{j} + 4\hat{k}) \cdot (10\hat{i} - 8\hat{j} - 4\hat{k})}{\sqrt{10^2 + (-8)^2 + (-4)^2}} \right| = \left| \frac{-60 - 32 - 16}{\sqrt{100 + 64 + 16}} \right| = \frac{108}{\sqrt{180}}$
Simplifying this expression yields $\frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}$.

Identifying the plane that contains two intersecting lines relies on the concept of a family of planes, where the equation $(4ax - y + 5z - 7a) + \lambda(2x - 5y - z - 3) = 0$ encompasses all planes passing through their intersection. Expanding this expression yields $(4a + 2\lambda)x - (1 + 5\lambda)y + (5 - \lambda)z - (7a + 3\lambda) = 0$. Since the line $\frac{x-4}{1} = \frac{y+1}{-2} = \frac{z}{1}$ lies within this plane, the point $(4, -1, 0)$ must be a solution to the plane's equation, and the normal vector of the plane must be orthogonal to the direction vector $(1, -2, 1)$.

Substituting the point $(4, -1, 0)$ into the plane equation results in $4(4a + 2\lambda) + (1 + 5\lambda) - (7a + 3\lambda) = 0$, which simplifies to $9a + 10\lambda + 1 = 0$. Furthermore, the dot product of the normal vector $(4a + 2\lambda, -(1 + 5\lambda), 5 - \lambda)$ and the direction vector $(1, -2, 1)$ being zero provides the equation $(4a + 2\lambda) - 2(-(1 + 5\lambda)) + (5 - \lambda) = 0$, leading to $4a + 11\lambda + 7 = 0$. Solving this system for $a$ and $\lambda$ yields $a = 1$ and $\lambda = -1$, which defines the specific plane $x + 2y + 3z - 2 = 0$.

To determine the intersection point $P(\alpha, \beta, \gamma)$, points on the line $\frac{x-3}{7} = \frac{y-2}{-1} = \frac{z-3}{-4}$ are represented as $(7t + 3, -t + 2, -4t + 3)$ for a parameter $t$. Substituting these coordinates into the plane equation $(7t + 3) + 2(-t + 2) + 3(-4t + 3) - 2 = 0$ results in $-7t + 14 = 0$, giving $t = 2$. Consequently, the coordinates of the point $P$ are $(17, 0, -5)$, and the sum $\alpha + \beta + \gamma$ is 12.

The reflection of a point $(x_1, y_1, z_1)$ across a plane defined by $Ax + By + Cz = D$ is located along the line passing through the point that is perpendicular to the plane, such that the midpoint of the segment connecting the point to its image lies on the plane. This relationship is expressed by the formula $\frac{a - x_1}{A} = \frac{b - y_1}{B} = \frac{c - z_1}{C} = \frac{-2(Ax_1 + By_1 + Cz_1 - D)}{A^2 + B^2 + C^2}$, where $(a, b, c)$ represents the coordinates of the mirror image.

Substituting the point $(2, 4, 7)$ and the plane $3x - y + 4z = 2$ into this equation, the value $Ax_1 + By_1 + Cz_1 - D$ becomes $3(2) - 1(4) + 4(7) - 2$, which evaluates to $28$. The sum of the squares of the normal vector components is $3^2 + (-1)^2 + 4^2 = 26$. Consequently, the proportionality constant for the ratios is $\frac{-2(28)}{26} = -\frac{28}{13}$.

Setting the coordinates to this ratio, the values for the image point $(a, b, c)$ are determined by $\frac{a - 2}{3} = -\frac{28}{13}$, $\frac{b - 4}{-1} = -\frac{28}{13}$, and $\frac{c - 7}{4} = -\frac{28}{13}$. Solving these equations yields $a = 2 - \frac{84}{13} = -\frac{58}{13}$, $b = 4 + \frac{28}{13} = \frac{80}{13}$, and $c = 7 - \frac{112}{13} = -\frac{21}{13}$. Evaluating the required expression $2a + b + 2c$ gives $2(-\frac{58}{13}) + \frac{80}{13} + 2(-\frac{21}{13}) = \frac{-116 + 80 - 42}{13} = \frac{-78}{13}$, resulting in $-6$.

A line lies on a plane if and only if any point on the line satisfies the plane equation and the line's direction vector is orthogonal to the plane's normal vector. The given line $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}$ passes through the point $(2, -1, -3)$ and has a direction vector $(3, -2, -1)$, while the plane $px - qy + z = 5$ has a normal vector $(p, -q, 1)$.

Substituting the point $(2, -1, -3)$ into the plane equation yields $p(2) - q(-1) + (-3) = 5$, which simplifies to the equation $2p + q = 8$. For the line to be contained within the plane, the dot product of the line's direction vector and the plane's normal vector must be zero, giving $(3)(p) + (-2)(-q) + (-1)(1) = 0$, which simplifies to $3p + 2q - 1 = 0$. Solving this system of linear equations by substituting $q = 8 - 2p$ into the second equation results in $3p + 2(8 - 2p) - 1 = 0$, leading to $p = 15$ and $q = -22$.