Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

Finding the intersection of a line and a plane requires treating the line as a set of parametric coordinates. By setting the ratios of the line equal to a parameter $\lambda$, any general point on the line is defined by $(2\lambda+4, 2\lambda+5, \lambda+3)$, and because this point must satisfy the plane equation $x+y+z=2$, substituting these coordinates yields $(2\lambda+4) + (2\lambda+5) + (\lambda+3) = 2$. This expression simplifies to $5\lambda + 12 = 2$, which results in $5\lambda = -10$ and $\lambda = -2$, confirming the intersection point is $(0, 1, 1)$. Testing these values in the equation $\frac{x-1}{1} = \frac{y-3}{2} = \frac{z+4}{-5}$ produces $-1$ for each ratio, which identifies this as the line passing through the point.

The angle between a line and a plane is determined by the dot product of the line's direction vector and the plane's normal vector using the sine function, defined as $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$. Given the direction ratios of the line as $(2, 1, -2)$ and the plane normal vector as $\vec{n} = (1, -2, -k)$, the magnitude of the line vector is $|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$, and the magnitude of the normal vector is $|\vec{n}| = \sqrt{1^2 + (-2)^2 + (-k)^2} = \sqrt{k^2 + 5}$. Calculating the dot product $\vec{b} \cdot \vec{n} = (2)(1) + (1)(-2) + (-2)(-k) = 2k$, the expression becomes $\sin \theta = \frac{|2k|}{3\sqrt{k^2 + 5}}$. Since the angle $\theta$ has $\cos \theta = \frac{2\sqrt{2}}{3}$, applying the identity $\sin^2 \theta + \cos^2 \theta = 1$ gives $\sin \theta = \sqrt{1 - \left(\frac{2\sqrt{2}}{3}\right)^2} = \sqrt{1 - \frac{8}{9}} = \frac{1}{3}$. Equating the two expressions for $\sin \theta$, we have $\frac{|2k|}{3\sqrt{k^2 + 5}} = \frac{1}{3}$, which simplifies to $\frac{|2k|}{\sqrt{k^2 + 5}} = 1$. Squaring both sides yields $\frac{4k^2}{k^2 + 5} = 1$, which leads to $3k^2 = 5$, resulting in $k = \sqrt{\frac{5}{3}}$.

To determine the dihedral angle between two faces of a tetrahedron, we find the angle between their respective normal vectors, which are calculated using the cross product of two vectors lying within each face. For the face OPQ, the normal vector $\vec{n_1}$ is found by the cross product of $\overset{\rightarrow}{OP} = (1, 2, 1)$ and $\overset{\rightarrow}{OQ} = (2, 1, 3)$, resulting in $\vec{n_1} = 5\hat{i} - \hat{j} - 3\hat{k}$. Similarly, for the face PQR, we calculate the normal $\vec{n_2}$ using the vectors $\overset{\rightarrow}{PQ} = (1, -1, 2)$ and $\overset{\rightarrow}{PR} = (-2, -1, 1)$, which yields $\vec{n_2} = \hat{i} - 5\hat{j} - 3\hat{k}$. The cosine of the angle $\theta$ between the faces is given by the formula $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. Substituting the values, we compute the dot product as $5(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$, and the magnitudes of the normal vectors are each $\sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{35}$. Thus, the final value is $\cos \theta = \frac{19}{35}$, corresponding to an angle of $\cos^{-1}\left(\frac{19}{35}\right)$.

The geometric interpretation of the reflection of a point across a plane provides the key to identifying the coordinates of point $P$. Since $Q(0, -1, -3)$ is the image of $P$ in the plane $3x - y + 4z - 2 = 0$, the line passing through $PQ$ must be normal to the plane, with direction ratios matching the plane's normal vector $(3, -1, 4)$. Using the reflection formula $\frac{x-0}{3} = \frac{y+1}{-1} = \frac{z+3}{4} = -2 \frac{3(0) - (-1) + 4(-3) - 2}{9+1+16}$, the expression simplifies to $1$, which establishes the coordinates of point $P$ as $(3, -2, 1)$.

With vertices at $P(3, -2, 1)$, $Q(0, -1, -3)$, and $R(3, -1, -2)$ identified, the area of $\Delta PQR$ is determined using the magnitude of the cross product of two vectors formed from these points. Constructing the vectors $\vec{PQ} = Q - P = (-3, 1, -4)$ and $\vec{PR} = R - P = (0, 1, -3)$, we compute the cross product:
$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 1 & -4 \\ 0 & 1 & -3 \end{vmatrix} = \mathbf{i}(1) - \mathbf{j}(9) + \mathbf{k}(-3) = (1, -9, -3)$
The magnitude of this resulting vector is $\sqrt{1^2 + (-9)^2 + (-3)^2} = \sqrt{91}$. Applying the area formula for a triangle, $\text{Area} = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$, the area is $\frac{\sqrt{91}}{2}$ square units.

Any plane passing through the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ can be represented by the equation $P_1 + \lambda P_2 = 0$. For the planes $x + y + z - 1 = 0$ and $2x + 3y - z + 4 = 0$, this family of planes is expressed as $(x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0$. Regrouping the terms, the equation becomes $(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$.

A plane is parallel to the $y$-axis if its normal vector is perpendicular to the $y$-axis direction vector, which is $(0, 1, 0)$. This geometric condition requires the coefficient of $y$ in the plane's equation to be zero. By setting $1 + 3\lambda = 0$, we find $\lambda = -1/3$. Substituting this parameter value into the family of equations, we get $(1 - 2/3)x + 0y + (1 + 1/3)z + (-4/3 - 1) = 0$, which simplifies to $\frac{1}{3}x + \frac{4}{3}z - \frac{7}{3} = 0$. Multiplying the entire equation by 3 yields the plane equation $x + 4z - 7 = 0$.

To identify which point lies on this plane, substitute the coordinates of the options into $x + 4z - 7 = 0$. Evaluating the point $(3, 2, 1)$ results in $3 + 4(1) - 7$, which equals $3 + 4 - 7 = 0$. Since this satisfies the equation, the plane passes through this point.

The geometry of this problem relies on finding a plane that contains both a given line and its projection on another plane, which requires the normal vector of our target plane to be perpendicular to both the direction vector of the line and the normal vector of the reference plane $2x + 3y - z = 5$. The direction vector of the line is $2\hat{i} - \hat{j} + 3\hat{k}$ and the normal of the reference plane is $2\hat{i} + 3\hat{j} - \hat{k}$, so the normal of the required plane is found by their cross product $(2\hat{i} - \hat{j} + 3\hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k}) = -8\hat{i} + 8\hat{j} + 8\hat{k}$, which simplifies to the direction ratio $(-1, 1, 1)$.

Since the line passes through $(3, -2, 1)$, substituting these coordinates and the normal vector into the standard point-normal form gives the equation:
$-1(x - 3) + 1(y + 2) + 1(z - 1) = 0$
This simplifies to the plane equation $-x + y + z + 4 = 0$. Testing the point $(2, 0, -2)$ by substituting its coordinates into this equation yields $-(2) + 0 + (-2) + 4 = 0$, confirming that this point lies on the required plane.

A plane is defined by a point it contains and a normal vector perpendicular to its surface. Because every line mentioned in the problem passes through the origin $(0, 0, 0)$, our required plane must also contain this point. The secondary plane containing the lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ has a normal vector found by calculating the cross product of their direction vectors, $(3, 4, 2)$ and $(4, 2, 3)$, which yields $8\hat{i} - \hat{j} - 10\hat{k}$.

The required plane must be perpendicular to this secondary plane and contain the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$. Therefore, its normal vector is determined by the cross product of the secondary plane's normal $8\hat{i} - \hat{j} - 10\hat{k}$ and the line's direction vector $2\hat{i} + 3\hat{j} + 4\hat{k}$. This cross product results in $26\hat{i} - 52\hat{j} + 26\hat{k}$. Applying this normal vector in the equation $Ax + By + Cz = 0$ provides $26x - 52y + 26z = 0$, which simplifies to $x - 2y + z = 0$.

Two lines in space are perpendicular if the sum of the products of their corresponding direction ratios equals zero, which is expressed as $l_1l_2 + m_1m_2 + n_1n_2 = 0$. The first line, defined by $x = ay + b$ and $z = cy + d$, can be rearranged into the symmetric form $\frac{x-b}{a} = \frac{y}{1} = \frac{z-d}{c}$, indicating that its direction ratios are $(a, 1, c)$. Similarly, the second line, given by $x = a'z + b'$ and $y = c'z + d'$, can be written as $\frac{x-b'}{a'} = \frac{y-d'}{c'} = \frac{z}{1}$, revealing that its direction ratios are $(a', c', 1)$. Multiplying these corresponding direction ratios and setting their sum to zero leads to the equation $a(a') + 1(c') + c(1) = 0$, which simplifies to $aa' + c + c' = 0$.

Any plane that perpendicularly bisects a line segment must pass through the segment's midpoint and have a normal vector parallel to the direction of the line segment itself. Calculating the midpoint of the points $(-3, -3, 4)$ and $(3, 7, 6)$ yields $(0, 2, 5)$, while the direction vector is found by taking the difference between the endpoints, which simplifies to $(3, 5, 1)$ after scaling. Applying the point-normal form of the plane equation gives $3(x - 0) + 5(y - 2) + 1(z - 5) = 0$, which simplifies to $3x + 5y + z = 15$. Substituting the coordinates $(4, 1, -2)$ into this equation yields $3(4) + 5(1) + (-2) = 15$, confirming that this point lies on the plane.

The distance between two parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ is found using the formula $d = \frac{|D_1-D_2|}{\sqrt{A^{2+}B^{2+}C^2}}$.

The plane $4x-2y+4z+\lambda=0$ can be rewritten as $2x-y+2z+\frac{\lambda}{2}=0$ to match the normal vector of $2x-y+2z+3=0$. Using the distance formula, we set $\frac{|3-\frac{\lambda}{2}|}{\sqrt{2^{2+}(-1)^{2+}2^2}} = \frac{1}{3}$, which simplifies to $\frac{|3-\frac{\lambda}{2}|}{3} = \frac{1}{3}$ and results in $|3-\frac{\lambda}{2}|=1$. Solving for $\lambda$ yields $\lambda=4$ or $\lambda=8$.

Similarly, for the plane $2x-y+2z+\mu=0$, the distance is $\frac{|3-\mu|}{\sqrt{2^{2+}(-1)^{2+}2^2}} = \frac{2}{3}$, which simplifies to $\frac{|3-\mu|}{3} = \frac{2}{3}$ or $|3-\mu|=2$. This equation yields $\mu=1$ or $\mu=5$. To maximize the sum $\lambda+\mu$, we select the larger values, $\lambda=8$ and $\mu=5$, which produce a maximum sum of 13.

Finding the intersection of two lines in three-dimensional space requires defining a general point on each line using parametric variables, such as $\lambda$ for the first line and $\mu$ for the second. Equating the $x$, $y$, and $z$ components generates a system of linear equations that, when solved, yields $\lambda = -1$ and $\mu = 1$. Substituting these values into either line equation reveals that the intersection point $R$ has coordinates $(2, -4, 7)$. Since the $xy$-plane acts as a mirror that preserves $x$ and $y$ values but flips the sign of the vertical coordinate, reflecting the point $R$ across this plane results in the coordinates $(2, -4, -7)$.

The most direct way to determine the perpendicular distance from a point to a line is to define an arbitrary point on that line using a scalar parameter $\lambda$. Given the line equation $\frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1}$, setting these expressions equal to $\lambda$ allows us to define any point $Q$ on the line as $(10\lambda - 3, -7\lambda + 2, \lambda)$. The vector connecting the fixed point $P(2, -1, 4)$ to this variable point $Q$ is found by subtracting the coordinates of $P$ from the coordinates of $Q$, resulting in the vector $(10\lambda - 5, -7\lambda + 3, \lambda - 4)$.

Since the line segment $PQ$ must be perpendicular to the line, the dot product of this vector and the direction ratios of the line, which are $(10, -7, 1)$, must equal zero. This relationship creates the equation $10(10\lambda - 5) - 7(-7\lambda + 3) + 1(\lambda - 4) = 0$. Solving this linear equation for $\lambda$ yields $\lambda = \frac{1}{2}$. Applying this value to the coordinates of $Q$ and calculating the magnitude of the resulting vector $PQ$ gives a distance of $\sqrt{\frac{50}{4}}$, which evaluates to approximately $3.54$. Because this calculated distance is greater than 3 but less than 4, it satisfies the provided criteria.

Four points are coplanar if and only if the scalar triple product of the vectors originating from one of the points to the other three is zero. By establishing the vectors $(1-\lambda^2, 2, 0)$, $(2, 1-\lambda^2, 0)$, and $(2, 2, -\lambda^{2-}1)$ relative to the point $(-1, -1, 1)$, we set their determinant to zero, which simplifies to $(\lambda^{2+}1)^2(3-\lambda^2) = 0$. Given that $\lambda$ must be a real number, the term $(\lambda^{2+}1)^2$ is strictly positive and cannot be zero, leaving $3-\lambda^2 = 0$ as the only condition to solve. Therefore, the set $S$ contains the values $\{\sqrt{3}, -\sqrt{3}\}$.

Since the plane contains two lines, its normal vector is determined by the cross product of their respective direction vectors, $(1, 2, -1)$ and $(-1, 1, -2)$. Calculating this product results in the vector $(-3, 3, 3)$, which simplifies to the direction ratios $(1, -1, -1)$ for the plane's normal. Using the common point $(1, 1, 0)$ shared by both lines, the equation of the plane is constructed as $1(x-1) - 1(y-1) - 1(z-0) = 0$, which simplifies to $x - y - z = 0$. The perpendicular distance from the point $(2, 1, 4)$ to this plane is found using the formula $\frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$. Substituting the given values yields $\frac{|2 - 1 - 4|}{\sqrt{1^2 + (-1)^2 + (-1)^2}}$, which simplifies to $\frac{|-3|}{\sqrt{3}}$ and produces the final distance of $\sqrt{3}$.

Any line in three-dimensional space can be parameterized by expressing the coordinates $x$, $y$, and $z$ in terms of a single variable $t$, allowing us to represent any arbitrary point on the line as $(3t+2, 2t-1, -t+1)$. To find the intersection points $P$ and $Q$ with the given planes, we substitute these parametric expressions directly into the equation of each plane. For the first plane, $2x+3y-z+13=0$, the substitution $2(3t+2)+3(2t-1)-(-t+1)+13=0$ simplifies to $13t+13=0$, which yields $t=-1$ and corresponds to the point $P=(-1,-3,2)$.

For the second plane, $3x+y+4z=16$, substituting the parametric coordinates gives $3(3t+2)+(2t-1)+4(-t+1)=16$, which simplifies to $7t+9=16$, resulting in $t=1$ and the point $Q=(5,1,0)$. The distance between these two points is then determined using the distance formula, $\sqrt{(5 - (-1))^2 + (1 - (-3))^2 + (0 - 2)^2}$, which simplifies to $\sqrt{6^2 + 4^2 + (-2)^2} = \sqrt{36+16+4} = \sqrt{56}$. This value further reduces to $2\sqrt{14}$.

A vector is parallel to a plane if and only if it is perpendicular to the plane's normal vector, which means their dot product must equal zero. For the plane defined by $x - 4y + 3z = 1$, the normal vector is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$. Subtracting the components of point A from point B yields the vector $\vec{AB} = (2 + 3\mu)\hat{i} + (3 - \mu)\hat{j} + (4 - 5\mu)\hat{k}$.

Applying the condition for perpendicularity, the dot product is calculated as $(2 + 3\mu)(1) + (3 - \mu)(-4) + (4 - 5\mu)(3) = 0$. Simplifying this expression leads to $2 + 3\mu - 12 + 4\mu + 12 - 15\mu = 0$. Combining the like terms results in $-8\mu + 2 = 0$, which simplifies to $\mu = \frac{1}{4}$.

Determining the plane that contains two lines relies on identifying a normal vector perpendicular to both lines, which is achieved by taking the cross product of their direction vectors $(3, 5, 7)$ and $(1, 4, 7)$. This calculation produces the vector $(7, -14, 7)$, which simplifies to the normal vector $(1, -2, 1)$. By utilizing this normal vector along with the point $(-2, 2, -5)$ found on the first line, the equation of the plane is determined to be $x - 2y + z + 11 = 0$. The perpendicular distance from the origin $(0, 0, 0)$ is then computed using the distance formula $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$, which yields $\frac{11}{\sqrt{1^2 + (-2)^2 + 1^2}}$ and results in $\frac{11}{\sqrt{6}}$.

Any point on the line defined by $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{1}$ can be expressed in terms of a parameter $t$ as $(2t+1, -t-1, t)$. Since the foot of the perpendicular $Q$ lies on the intersection of the planes $x+y+z=3$ and $x-y+z=3$, we can determine its coordinates by solving this system. Subtracting the second equation from the first yields $2y=0$, so $y=0$, and adding them results in $2x+2z=6$, or $z=3-x$. Thus, any point $Q$ on the intersection line takes the form $(s, 0, 3-s)$ for some parameter $s$.

The vector $\vec{PQ}$ represents the perpendicular segment from the line to the plane $x+y+z=3$, so it must be parallel to the plane's normal vector $\vec{n} = (1, 1, 1)$. Subtracting the coordinates of $P$ from $Q$ gives the vector $(s-2t-1, t+1, 3-s-t)$. For this vector to be parallel to $(1, 1, 1)$, all its components must be equal to one another. Setting the first component equal to the second, $s-2t-1 = t+1$, simplifies to $s-3t=2$. Setting the second equal to the third, $t+1 = 3-s-t$, simplifies to $2t+s=2$. Solving this system of linear equations yields $t=0$ and $s=2$. Substituting $s=2$ back into the coordinates for $Q$ results in $(2, 0, 1)$.

A plane parallel to two given lines has a normal vector orthogonal to their direction vectors, $(3, -1, 2)$ and $(1, 2, 3)$, which we find using the cross product $\vec{n} = \hat{i}(-3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - (-1)) = -7\hat{i} - 7\hat{j} + 7\hat{k}$. Simplifying this normal vector by dividing by $-7$ gives the direction ratios $(1, 1, -1)$. We then apply the point-normal form $1(x - 4) + 1(y + 1) - 1(z - 2) = 0$, which simplifies to the plane equation $x + y - z = 1$. Substituting the coordinates from the final option, $(1, 1, 1)$, confirms that $1 + 1 - 1 = 1$, indicating the plane passes through this specific point.

The vector perpendicular to the plane containing the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$ is their cross product, which results in $\hat{i}-2\hat{j}+\hat{k}$. Calculating the magnitude of the projection involves finding the absolute value of the dot product of $2\hat{i}+3\hat{j}+\hat{k}$ with this perpendicular vector and dividing by the magnitude of that perpendicular vector. The dot product is $(2)(1) + (3)(-2) + (1)(1) = -3$, and the magnitude of the perpendicular vector is $\sqrt{1^{2+}(-2)^{2+}1^2} = \sqrt{6}$. Dividing the absolute value of the dot product, 3, by the magnitude $\sqrt{6}$ yields $\sqrt{\frac{3}{2}}$.