Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

A plane intersecting the coordinate axes at points $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$ is expressed by the intercept form equation $\frac{X}{a} + \frac{Y}{b} + \frac{Z}{c} = 1$. The perpendicular distance $d$ from the origin $(0, 0, 0)$ to this plane is given by $d = \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)^{-1/2}$. Setting this distance equal to 3 leads to the relation $\left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)^{-1/2} = 3$, which simplifies to $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9}$.

The centroid $(x, y, z)$ of $\triangle ABC$ is defined by the coordinates $x = \frac{a+0+0}{3}$, $y = \frac{0+b+0}{3}$, and $z = \frac{0+0+c}{3}$, which gives $a = 3x$, $b = 3y$, and $c = 3z$. Substituting these expressions into the established distance equation yields $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{9}$, simplifying to $\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{9}$. Multiplying this entire equation by 9 results in the final locus $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1$.

A plane containing two lines has a normal vector perpendicular to both, which is determined by the cross product of their direction vectors. Given the direction vectors $(6, 7, 8)$ and $(3, 5, 7)$, the cross product yields $(9, -18, 9)$, which simplifies to the normal vector $(1, -2, 1)$. By utilizing a point from one of the lines, such as $(-1, 1, 3)$, the plane equation can be constructed as $1(x+1) - 2(y-1) + 1(z-3) = 0$, which simplifies to $x - 2y + z = 0$.

The foot of the perpendicular from $(1, -2, 1)$ to this plane lies on a line passing through the point with direction ratios defined by the plane's normal vector. This line is expressed in parametric form as $\frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-1}{1} = k$, where $k$ is a scalar. Substituting the parametric expressions $x = k+1$, $y = -2k-2$, and $z = k+1$ into the plane equation $x - 2y + z = 0$ results in $(k+1) - 2(-2k-2) + (k+1) = 0$. This simplifies to $6k + 6 = 0$, or $k = -1$, which identifies the foot of the perpendicular as $(0, 0, 0)$.

The condition for a line to lie in a plane requires the line's point $(3, -2, -\lambda)$ to satisfy $2x - 4y + 3z = 2$ and its direction vector $(1, -1, -2)$ to be orthogonal to the plane's normal $(2, -4, 3)$. Substituting the point gives $2(3) - 4(-2) + 3(-\lambda) = 2$, which simplifies to $14 - 3\lambda = 2$ and results in $\lambda = 4$. With $\lambda = 4$, the first line is expressed as the parametric set $(k+3, -k-2, -2k-4)$, while the second line is represented by $(12t+1, 9t, 4t)$. Equating the coordinates leads to $k+3 = 12t+1$ and $-k-2 = 9t$, which simplify to $k = 12t-2$ and, upon substitution into the second equation, $-(12t-2)-2 = 9t$, resulting in $t = 0$. Since both lines pass through the point $(1, 0, 0)$ when $t = 0$ and $k = -2$, the lines intersect, making the shortest distance $0$.

For a homogeneous system of linear equations $AX=0$ to have non-trivial solutions, the determinant of the coefficient matrix must be zero. The determinant of the coefficient matrix is:
$\begin{vmatrix} 1 & 8 & 7 \\ 9 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix} = 1(2-3) - 8(9-3) + 7(9-2) = -1 - 48 + 49 = 0$
Since the determinant is zero, the system has non-trivial solutions. Expressing $x$ and $y$ in terms of $z$ using $x+y+z=0$ and $x+8y+7z=0$ yields $y = -\frac{6}{7}z$ and $x = -\frac{1}{7}z$. Substituting these values into the plane equation $x+2y+z=6$ results in:
$-\frac{z}{7} + 2\left(-\frac{6}{7}z\right) + z = 6 \implies -\frac{6}{7}z = 6 \implies z = -7$
With $z = -7$, the values are $a = 1$, $b = 6$, and $c = -7$, thus $2a+b+c = 2(1) + 6 - 7 = 1$.

Two lines in three-dimensional space are coplanar if the scalar triple product of their direction vectors and the vector connecting a point on each line is zero. With points $(1, 2, -3)$ and $(3, 2, 1)$ on the lines, the connecting vector is $(2, 0, 4)$, creating the following determinant condition for coplanarity:

$\begin{vmatrix} 1 & 2 & \lambda^2 \\ 1 & \lambda^2 & 2 \\ 2 & 0 & 4 \end{vmatrix} = 0$

Evaluating this expression results in $4\lambda^2 - 2\lambda^4 = 0$, which factors into $2\lambda^2(2 - \lambda^2) = 0$. This equation is satisfied when $\lambda = 0$ or $\lambda = \pm\sqrt{2}$, yielding three distinct real values for $\lambda$.

For a straight line to lie entirely within a plane, two geometric conditions must be satisfied simultaneously. First, the line's direction vector must be perpendicular to the normal vector of the plane, which means their dot product is zero. The direction vector of the line given by $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$ is $(2, -1, 3)$, and the normal vector to the plane $lx + my - z = 9$ is $(l, m, -1)$. Calculating their dot product results in the equation:
$2l - m - 3 = 0$

Second, every point on the line must also satisfy the equation of the plane. By substituting the point $(3, -2, -4)$, which clearly lies on the line, into the plane equation $lx + my - z = 9$, we obtain:
$l(3) + m(-2) - (-4) = 9$
$3l - 2m + 4 = 9$
$3l - 2m = 5$

With these two equations, $2l - m = 3$ and $3l - 2m = 5$, we can solve for the unknowns. Multiplying the first equation by $2$ gives $4l - 2m = 6$. Subtracting the second equation, $3l - 2m = 5$, from this new equation yields $l = 1$. Substituting $l = 1$ back into the first equation results in $2(1) - m = 3$, which simplifies to $m = -1$. The value of $l^2 + m^2$ is therefore $1^2 + (-1)^2$, which equals 2.

A vector is considered equally inclined to the coordinate axes when its direction ratios are equal in magnitude. This geometric property implies that for a vector $\vec{v} = (a, b, c)$, the condition $|a| = |b| = |c|$ must be satisfied. First, identify the midpoint $D$ of the line segment $BC$ by averaging the coordinates of $B(-1, 3, 2)$ and $C(\lambda, 5, \mu)$, which gives $D\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right)$. The direction vector $\vec{AD}$ is then determined by subtracting the coordinates of $A(2, 3, 5)$ from $D$, resulting in the vector $\left(\frac{\lambda-5}{2}, 1, \frac{\mu-8}{2}\right)$. Setting the magnitude of each component equal to the magnitude of the known y-component, which is 1, provides the equations $\left|\frac{\lambda-5}{2}\right| = 1$ and $\left|\frac{\mu-8}{2}\right| = 1$. Solving for the values gives $\frac{\lambda-5}{2} = 1$, leading to $\lambda = 7$, and $\frac{\mu-8}{2} = 1$, leading to $\mu = 10$. Finally, substituting these values into the expression $\lambda^3 + \mu^3 + 5$ results in $7^3 + 10^3 + 5$, which simplifies to $343 + 1000 + 5$, totaling 1348.

The distance of a point from a plane measured along a specific line is calculated by dividing the perpendicular distance from the point to the plane by the cosine of the angle between the line's direction and the normal vector of the plane. The perpendicular distance from the point $(1, -5, 9)$ to the plane $x-y+z-5=0$ is given by the formula $\frac{|1 - (-5) + 9 - 5|}{\sqrt{1^2 + (-1)^2 + 1^2}}$, which simplifies to $\frac{10}{\sqrt{3}}$. The normal vector to the plane is $\vec{n} = (1, -1, 1)$ and the direction vector of the line $x=y=z$ is $\vec{v} = (1, 1, 1)$, making the cosine of the angle $\theta$ between them $\cos \theta = \frac{(1)(1) + (-1)(1) + (1)(1)}{\sqrt{3}\sqrt{3}} = \frac{1}{3}$. Dividing the perpendicular distance by $\cos \theta$ results in the total distance along the line, yielding $10\sqrt{3}$.

To determine the equation of a plane that is perpendicular to two other planes, we first find its normal vector by calculating the cross product of the normal vectors of the given planes, which are $(1, -1, 2)$ and $(2, -2, 1)$. The cross product of these vectors is

, which gives us a normal vector direction of $(1, 1, 0)$.

Since the plane passes through the point $(1, 2, 2)$, we can write its equation using the point-normal form $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$, leading to $1(x-1) + 1(y-2) + 0(z-2) = 0$. This simplifies to the plane equation $x + y - 3 = 0$. We then calculate the distance from the point $(1, -2, 4)$ to this plane using the standard perpendicular distance formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$. Substituting our values, we get $d = \frac{|1(1) + 1(-2) - 3|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 - 2 - 3|}{\sqrt{2}} = \frac{|-4|}{\sqrt{2}}$, which simplifies to $2\sqrt{2}$.

The shortest distance between two skew lines is determined by the projection of the vector connecting an arbitrary point on each line onto the common perpendicular of their direction vectors. Using the formula for the shortest distance $d$ between lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$:
$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$
From the given equations, the first line passes through the origin $\vec{a_1} = 0\hat{i} + 0\hat{j} + 0\hat{k}$ with direction $\vec{b_1} = 2\hat{i} + 2\hat{j} + 1\hat{k}$, while the second line passes through $\vec{a_2} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ with direction $\vec{b_2} = -1\hat{i} + 8\hat{j} + 4\hat{k}$. The vector joining the two points is $\vec{a_2} - \vec{a_1} = -2\hat{i} + 4\hat{j} + 5\hat{k}$, and the cross product of the direction vectors is $\vec{b_1} \times \vec{b_2} = (2\hat{i} + 2\hat{j} + 1\hat{k}) \times (-1\hat{i} + 8\hat{j} + 4\hat{k}) = 0\hat{i} - 9\hat{j} + 18\hat{k}$. Calculating the shortest distance yields the scalar projection $\frac{|(-2\hat{i} + 4\hat{j} + 5\hat{k}) \cdot (0\hat{i} - 9\hat{j} + 18\hat{k})|}{\sqrt{0^2 + (-9)^2 + 18^2}} = \frac{|0 - 36 + 90|}{\sqrt{405}} = \frac{54}{9\sqrt{5}} = \frac{6}{\sqrt{5}}$. Since $6/\sqrt{5}$ is approximately $2.68$, this value falls within the interval $(2, 3]$.

Any plane passing through the line of intersection of two given planes $P_1 = 0$ and $P_2 = 0$ can be represented by the family of planes equation $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar constant. Because the required plane must be parallel to an existing plane, the normal vectors of the two planes must be proportional to one another, which allows us to equate the ratios of their respective coefficients to solve for the unknown parameter $\lambda$.

Starting with the given planes $2x-5y+z-3=0$ and $x+y+4z-5=0$, the general equation for the plane containing their line of intersection is $(2x-5y+z-3) + \lambda(x+y+4z-5) = 0$. Grouping the terms by variables, we can rewrite this as $(2+\lambda)x + (\lambda-5)y + (1+4\lambda)z - (3+5\lambda) = 0$. Since this plane must be parallel to $x+3y+6z-1=0$, the coefficients of $x, y,$ and $z$ in both equations must be in the same proportion, leading to the condition $\frac{2+\lambda}{1} = \frac{\lambda-5}{3} = \frac{1+4\lambda}{6}$.

Solving the first part of this equality, $3(2+\lambda) = \lambda-5$, gives $6+3\lambda = \lambda-5$, which simplifies to $2\lambda = -11$, or $\lambda = -11/2$. Substituting this value back into the general family equation results in $(2x-5y+z-3) - \frac{11}{2}(x+y+4z-5) = 0$. Multiplying the entire expression by 2 to eliminate the denominator yields $4x-10y+2z-6 - 11x - 11y - 44z + 55 = 0$. Combining like terms, we obtain $-7x-21y-42z+49=0$, and dividing by $-7$ provides the simplified form $x+3y+6z-7=0$, which rearranges to $x+3y+6z=7$.

To find the intersection of a line and a plane, represent any point on the line using a parameter $\lambda$ such that $x = 3\lambda + 2$, $y = 4\lambda - 1$, and $z = 12\lambda + 2$. Substituting these coordinates into the plane equation $x - y + z = 16$ gives $(3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 16$, which simplifies to $11\lambda + 5 = 16$, revealing that $\lambda = 1$ at the point of intersection.

Applying $\lambda = 1$ to the parametric line equations identifies the intersection point as $(5, 3, 14)$. The distance between this point and the given coordinate $(1, 0, 2)$ is found using the 3D distance formula, yielding $\sqrt{(5 - 1)^2 + (3 - 0)^2 + (14 - 2)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$.

The angle between two lines defined by their direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is determined by the expression $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$, where each set of direction cosines satisfies the fundamental constraint $l^{2+}m^{2+}n^2=1$.

Substituting $l = -(m+n)$ from the first equation into the second equation $l^2 = m^{2+}n^2$ leads to $(m+n)^2 = m^{2+}n^2$, which simplifies to $2mn=0$, indicating that either $m=0$ or $n=0$. Choosing $m=0$ requires $l=-n$, and applying the normalization condition $l^{2+}m^{2+}n^2=1$ results in $(-n)^{2+}0^{2+}n^2=1$, giving the direction cosines $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$. Similarly, setting $n=0$ leads to $l=-m$ and produces the direction cosines $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$. Computing the cosine of the angle between these two lines yields $\cos \theta = |(-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) + (0)(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(0)| = \frac{1}{2}$, which identifies the angle as $\frac{\pi}{3}$.

Reflecting a line parallel to a plane results in a new line that maintains the same direction ratios while passing through the image of any arbitrary point from the original line. Verifying the relationship confirms that the line's direction vector $(3, 1, -5)$ and the plane's normal vector $(2, -1, 1)$ yield a dot product of $(3)(2) + (1)(-1) + (-5)(1) = 0$, proving the line is parallel to the plane. The reflection of the point $(1, 3, 4)$ across the plane $2x - y + z + 3 = 0$ is the point $(-3, 5, 2)$. Using this image point and the original direction vector $(3, 1, -5)$ produces the equation $\frac{x+3}{3} = \frac{y-5}{1} = \frac{z-2}{-5}$.

Parallel planes share the same normal vector, so their perpendicular distance is calculated by the formula $\frac{|d_1 - d_2|}{\sqrt{A^2 + B^2 + C^2}}$, where $d_1$ and $d_2$ represent the constant terms when the equations are normalized to have identical coefficients. Multiplying the first plane's equation, $2x + y + 2z = 8$, by 2 produces $4x + 2y + 4z - 16 = 0$. Comparing this with the second plane, $4x + 2y + 4z + 5 = 0$, reveals the constant terms are $-16$ and $5$. Substituting these values into the distance formula results in $\frac{|-16 - 5|}{\sqrt{4^2 + 2^2 + 4^2}}$, leading to $\frac{|-21|}{\sqrt{16 + 4 + 16}}$, which evaluates to $\frac{21}{\sqrt{36}}$ or $\frac{21}{6}$ and simplifies to $\frac{7}{2}$.

Two lines in three-dimensional space are coplanar if and only if the scalar triple product of the displacement vector connecting a point from each line and the two direction vectors equals zero. By identifying the given points as $(2, 3, 4)$ and $(1, 4, 5)$ and the direction vectors as $(1, 1, -k)$ and $(k, 2, 1)$, the condition for coplanarity is established by the determinant equation

. Expanding this determinant along the first row leads to the quadratic equation $k^{2+}3k=0$, which factors as $k(k+3)=0$. Since this quadratic equation provides two distinct roots for $k$, there are exactly two values that satisfy the condition.

Two lines in three-dimensional space intersect if and only if the shortest distance between them is zero, which is verified by setting the scalar triple product of the relative position vector and the two direction vectors equal to zero. This geometric constraint ensures the lines are coplanar and not parallel. For the lines provided, the first line contains the point $(1, -1, 1)$ with direction $(2, 3, 4)$, and the second line contains the point $(3, k, 0)$ with direction $(1, 2, 1)$.

The intersection condition is satisfied when the determinant expression formed by the difference in position vectors and the direction vectors equals zero, represented as

. Simplifying the first row yields the determinant

. Expanding this along the top row gives $2(3 - 8) - (k+1)(2 - 4) - 1(4 - 3) = 0$, which simplifies to $-10 + 2(k+1) - 1 = 0$. Solving the resulting linear equation $2k - 9 = 0$ provides the value $k = \frac{9}{2}$.

Any plane parallel to $x-2y+2z-5=0$ must share the same coefficients for the variable terms, meaning it takes the general form $x-2y+2z+d=0$. The perpendicular distance from the origin $(0,0,0)$ to a plane represented by $Ax+By+Cz+d=0$ is determined by the formula $\frac{|d|}{\sqrt{A^{2+}B^{2+}C^2}}$. Substituting the coefficients from the parallel plane, the denominator becomes $\sqrt{1^{2+}(-2)^{2+}2^2}$, which simplifies to $\sqrt{1+4+4}$, or $3$.

Given that the distance from the origin is $1$, we set $\frac{|d|}{3}=1$, which results in $|d|=3$. This implies that the constant $d$ can be either $3$ or $-3$, leading to the two possible equations $x-2y+2z+3=0$ and $x-2y+2z-3=0$. The equation $x-2y+2z-3=0$ matches one of the provided options.

The geometric principle of a mirror image in a 3D line requires that the line acts as a perpendicular bisector of the segment connecting the point and its reflected image. This entails two distinct conditions: the line must be perpendicular to the segment, and the midpoint of that segment must lie on the line.

The direction ratios of the segment connecting $A(1,0,7)$ and $B(1,6,3)$ are $(0, 6, -4)$. The dot product of these ratios with the direction ratios $(1, 2, 3)$ of the line results in $1(0) + 2(6) + 3(-4) = 0$, confirming the line is orthogonal to the segment. Furthermore, the midpoint of segment $AB$ is located at $(\frac{1+1}{2}, \frac{0+6}{2}, \frac{7+3}{2}) = (1, 3, 5)$. Substituting these coordinates into the line equation $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ yields $\frac{1}{1} = \frac{3-1}{2} = \frac{5-2}{3} = 1$, which verifies that the midpoint lies exactly on the line.

Because both the perpendicularity and bisection conditions hold, point $A$ is indeed the mirror image of $B$. Statement $-2$ correctly observes that the line bisects the segment, but because bisection is only one of the necessary geometric criteria, it is not a sufficient explanation for why the line functions as a mirror image. While both statements are individually true, the second does not provide the comprehensive reasoning required to establish the first.

The angle between a line and a plane is determined by the relationship between the line's direction vector $\vec{b}$ and the plane's normal vector $\vec{n}$ using the sine function. Specifically, if $\theta$ is the angle between the line and the plane, then

$\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$

Given the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda}$, the direction vector is $\vec{b} = \langle 1, 2, \lambda \rangle$. For the plane $x+2y+3z=4$, the normal vector is $\vec{n} = \langle 1, 2, 3 \rangle$.

Calculating the components of the sine formula, the dot product is $1(1) + 2(2) + \lambda(3) = 5 + 3\lambda$, and the magnitudes of the vectors are $|\vec{b}| = \sqrt{1^2 + 2^2 + \lambda^2} = \sqrt{5 + \lambda^2}$ and $|\vec{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$. Since the problem provides the angle $\theta = \cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$, we use the identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$ to find that $\sin \theta = \sqrt{1 - \frac{5}{14}} = \frac{3}{\sqrt{14}}$.

Equating these expressions yields

$\frac{5 + 3\lambda}{\sqrt{14}\sqrt{5 + \lambda^2}} = \frac{3}{\sqrt{14}}$

Canceling the $\sqrt{14}$ from both denominators simplifies the equation to $5 + 3\lambda = 3\sqrt{5 + \lambda^2}$. Squaring both sides results in $(5 + 3\lambda)^2 = 9(5 + \lambda^2)$, which expands to $25 + 30\lambda + 9\lambda^2 = 45 + 9\lambda^2$. Subtracting $9\lambda^2$ from both sides gives $30\lambda = 20$, and solving for $\lambda$ produces $\lambda = \frac{2}{3}$.