Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The orientation of any line in three-dimensional space is governed by the identity $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$, where $\alpha$, $\beta$, and $\gamma$ represent the angles the line makes with the coordinate axes. Substituting the given values of $45^{\circ}$ and $120^{\circ}$ for the $x$ and $y$ axes results in the equation $\cos^2 45^{\circ} + \cos^2 120^{\circ} + \cos^2 \theta = 1$, which simplifies to $\frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 1$. This calculation leaves $\cos^2 \theta = \frac{1}{4}$, and because the angle $\theta$ with the $z$-axis is acute, we identify $\cos \theta = \frac{1}{2}$, which corresponds to $\theta = 60^{\circ}$.

A point is considered the mirror image of another with respect to a plane only when two geometric criteria are simultaneously fulfilled. First, the midpoint of the line segment connecting the two points must lie on the plane, indicating that the points are equidistant from it. Second, the line segment joining these points must be perpendicular to the plane, ensuring the path between them follows the normal vector.

We can test the validity of the second statement by finding the midpoint of $A(3,1,6)$ and $B(1,3,4)$, which is $(\frac{3+1}{2}, \frac{1+3}{2}, \frac{6+4}{2}) = (2,2,5)$. Substituting these coordinates into the equation $x-y+z=5$ results in $2-2+5=5$, which confirms the midpoint indeed lies on the plane. This verifies that Statement-2 is true.

To determine if the points form a mirror image, we examine the direction ratios of the segment $AB$, which are $(3-1, 1-3, 6-4) = (2, -2, 2)$. Comparing these to the normal vector of the plane, which is $(1, -1, 1)$, we observe that $(2, -2, 2) = 2(1, -1, 1)$. Because the line segment $AB$ is parallel to the normal vector, it is perpendicular to the plane, fulfilling the second necessary condition for reflection. Consequently, Statement-1 is also true. While Statement-2 correctly identifies that the plane bisects the segment, this condition alone does not establish perpendicularity, meaning it does not fully explain why $A$ is the mirror image of $B$.

A line is contained entirely within a plane if every point on the line satisfies the plane's equation and the line's direction vector is orthogonal to the plane's normal vector. From the given line $\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$, we can identify that the line passes through the point $(2, 1, -2)$ and has a direction vector of $(3, -5, 2)$.

Because the line lies within the plane $x+3y-\alpha z+\beta=0$, the plane's normal vector, which is $(1, 3, -\alpha)$, must be perpendicular to the line's direction. The dot product of these two vectors must equal zero, so we calculate $3(1) + (-5)(3) + 2(-\alpha) = 0$. This simplifies to $3 - 15 - 2\alpha = 0$, which yields $\alpha = -6$.

Since the plane must contain the line, the point $(2, 1, -2)$ must also satisfy the plane equation. Substituting the coordinates of this point and the calculated value of $\alpha$ into the equation $x+3y-\alpha z+\beta=0$, we get $2 + 3(1) - (-6)(-2) + \beta = 0$. Simplifying this arithmetic gives $2 + 3 - 12 + \beta = 0$, which results in $\beta = 7$, and thus the final values are $(\alpha, \beta) = (-6, 7)$.

The projections of a vector onto the three coordinate axes serve as its direction ratios, describing the vector's orientation in three-dimensional space. To determine the direction cosines, you must normalize these ratios by dividing each component by the total magnitude of the vector.

Calculate the magnitude of the vector by taking the square root of the sum of the squares of its components, resulting in $\sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. By dividing the direction ratios $6, -3,$ and $2$ by this calculated magnitude of $7$, you arrive at the direction cosines $\frac{6}{7}, \frac{-3}{7}, \frac{2}{7}$.

Any line in three-dimensional space determined by two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ can be expressed using a parameter $\lambda$ such that the coordinates satisfy the ratios of their differences. For the given points $(5, 1, a)$ and $(3, b, 1)$, this relationship is written as $\frac{x-5}{3-5} = \frac{y-1}{b-1} = \frac{z-a}{1-a} = \lambda$, which simplifies to $\frac{x-5}{-2} = \frac{y-1}{b-1} = \frac{z-a}{1-a} = \lambda$. Because the line intersects the $yz$-plane where the $x$-coordinate must be zero, we set the $x$ component of the parametric form to zero, meaning $-2\lambda + 5 = 0$, which gives the parameter value $\lambda = \frac{5}{2}$.

Substituting this value of $\lambda$ into the parametric expressions for the $y$ and $z$ coordinates allows us to match them with the intersection point $\left(0, \frac{17}{2}, -\frac{13}{2}\right)$. For the $y$-coordinate, the equation $(b-1)\left(\frac{5}{2}\right) + 1 = \frac{17}{2}$ simplifies to $\frac{5}{2}(b-1) = \frac{15}{2}$, resulting in $b-1 = 3$, or $b = 4$. Similarly, for the $z$-coordinate, the equation $(1-a)\left(\frac{5}{2}\right) + a = -\frac{13}{2}$ simplifies to $\frac{5}{2} - \frac{5}{2}a + a = -\frac{13}{2}$, which rearranges to $-\frac{3}{2}a = -9$, confirming that $a = 6$.

Two straight lines in three-dimensional space intersect if they are coplanar, which mathematically requires the scalar triple product of the difference of their position vectors and their direction vectors to be zero. For the given lines $\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$, the position vectors are $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{a}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$, while the direction vectors are $\vec{b}_1 = k\hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b}_2 = 3\hat{i} + k\hat{j} + 2\hat{k}$. The difference between the position vectors is $\vec{a}_2 - \vec{a}_1 = \hat{i} + \hat{j} - 2\hat{k}$. The intersection condition implies that the determinant formed by these vectors must equal zero:

$\begin{vmatrix} 1 & 1 & -2 \\ k & 2 & 3 \\ 3 & k & 2 \end{vmatrix} = 0$

Expanding this determinant results in $1(4-3k) - 1(2k-9) - 2(k^{2-}6) = 0$, which simplifies to the quadratic equation $-2k^2 - 5k + 25 = 0$. Solving this equation yields $k = \frac{5}{2}$ or $k = -5$, and given the requirement that $k$ must be an integer, the value is $k = -5$.

Every line in space relates to the coordinate axes through direction cosines, which are the cosines of the angles formed with each axis. The defining principle for these values is that the sum of the squares of the direction cosines, $l^2 + m^2 + n^2 = 1$, must always equal unity. Given that the line makes an angle of $\frac{\pi}{4}$ with both the $x$ and $y$ axes, the corresponding values are $l = \cos(\frac{\pi}{4})$ and $m = \cos(\frac{\pi}{4})$. Substituting the squares of these values into the identity yields $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 + n^2 = 1$. This simplifies to $\frac{1}{2} + \frac{1}{2} + n^2 = 1$, resulting in $1 + n^2 = 1$, or $n^2 = 0$. Consequently, the cosine of the angle with the $z$-axis is zero, which signifies that the line is perpendicular to the $z$-axis, forming an angle of $\frac{\pi}{2}$.

The direction of a line formed by the intersection of two planes is perpendicular to the normal vectors of both planes. For the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$, these normal vectors are $(2, 3, 1)$ and $(1, 3, 2)$. To find the direction vector $(l, m, n)$ of the line, we can solve the simultaneous equations $2l + 3m + n = 0$ and $l + 3m + 2n = 0$ using cross-multiplication.

Solving these yields the ratios $\frac{l}{6-3} = \frac{m}{1-4} = \frac{n}{6-3}$, which simplifies to $\frac{l}{3} = \frac{m}{-3} = \frac{n}{3}$. Dividing by 3, we obtain the proportional direction vector $(1, -1, 1)$. To convert this into direction cosines, we divide each component by the magnitude of the vector, which is $\sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$. This gives $l = \frac{1}{\sqrt{3}}$, $m = -\frac{1}{\sqrt{3}}$, and $n = \frac{1}{\sqrt{3}}$. Since the direction cosine $l$ represents $\cos \alpha$ for the angle $\alpha$ formed with the positive $x$-axis, it follows that $\cos \alpha = \frac{1}{\sqrt{3}}$.

The center of a sphere is always the midpoint of any diameter connecting two points on its surface. By extracting the center from the given sphere equation $x^{2+}y^{2+}z^{2-}6x-12y-2z+20=0$, we find the center to be $(3,6,1)$. Applying the midpoint formula between the known end $(2,3,5)$ and the unknown end $(x,y,z)$, we establish the relations $\frac{x+2}{2}=3$, $\frac{y+3}{2}=6$, and $\frac{z+5}{2}=1$. Solving these equations yields $x=4$, $y=9$, and $z=-3$, identifying the coordinates of the other end as $(4,9,-3)$.

The reflection of a point across a plane creates a situation where the plane acts as the perpendicular bisector of the line segment connecting the original point to its image. To determine the reflected point, define a line passing through $(-1, 3, 4)$ with direction ratios matching the plane's normal, $(1, -2, 0)$, expressed as $\frac{x+1}{1} = \frac{y-3}{-2} = \frac{z-4}{0} = k$. Substituting the general coordinates $(k-1, -2k+3, 4)$ into the plane equation $x - 2y = 0$ leads to $(k-1) - 2(-2k+3) = 0$, which simplifies to $5k - 7 = 0$ or $k = \frac{7}{5}$.

This value of $k$ identifies the foot of the perpendicular as $N\left(\frac{2}{5}, \frac{1}{5}, 4\right)$. Treating $N$ as the midpoint between the original point and the unknown image coordinates $(\alpha, \beta, \gamma)$ requires solving $\frac{\alpha-1}{2} = \frac{2}{5}$, $\frac{\beta+3}{2} = \frac{1}{5}$, and $\frac{\gamma+4}{2} = 4$. These calculations yield an image point of $\left(\frac{9}{5}, -\frac{13}{5}, 4\right)$, a result that does not appear among the listed options.

Two lines are perpendicular when the dot product of their direction vectors is zero, which means the sum of the products of their corresponding direction ratios must equal zero. By rearranging the provided equations into the standard symmetric form $\frac{x-b}{a} = \frac{y}{1} = \frac{z-d}{c}$ and $\frac{x-b'}{a'} = \frac{y}{1} = \frac{z-d'}{c'}$, we can identify the direction vectors as $(a, 1, c)$ and $(a', 1, c')$. Applying the condition for perpendicularity gives $(a)(a') + (1)(1) + (c)(c') = 0$, which simplifies directly to $aa' + cc' = -1$.

The intersection of a plane and a sphere creates a circle where the sphere's radius, the circle's radius, and the perpendicular distance from the sphere's center to the plane form a right triangle. By rearranging the sphere equation $x^{2+}y^{2+}z^{2-}x+z-2=0$ into the standard form $\left( x-\frac{1}{2} \right)^{2+}y^{2+}\left( z+\frac{1}{2} \right)^2=\frac{5}{2}$, we determine that the center is located at $\left( \frac{1}{2}, 0, -\frac{1}{2} \right)$ and the square of the radius is $\frac{5}{2}$. The perpendicular distance from this center to the plane $x+2y-z-4=0$ is given by
$d = \frac{ \left| \frac{1}{2} + 2(0) - \left( -\frac{1}{2} \right) - 4 \right| }{ \sqrt{1^2 + 2^2 + (-1)^2} } = \frac{|1-4|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}}$
Using the Pythagorean theorem to relate these values, the square of the circle's radius is the difference between the square of the sphere's radius and the squared perpendicular distance, expressed as $\frac{5}{2} - \frac{3}{2} = 1$. Consequently, the radius of the circle is 1.

Identifying the centers of the two spheres is the first step, derived from their respective equations to obtain $(-3, 4, 1)$ and $(5, -2, 1)$. Calculating the midpoint of the segment joining these centers yields $(1, 1, 1)$, a coordinate that must lie on the specified plane. Substituting these values into the plane equation $2ax - 3ay + 4az + 6 = 0$ results in $2a(1) - 3a(1) + 4a(1) + 6 = 0$. Simplifying this expression leads to $3a + 6 = 0$, which determines the value of $a$ to be $-2$.

The angle $\theta$ between a line and a plane is determined by the relationship $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$, where $\vec{b}$ is the direction vector of the line and $\vec{n}$ is the normal vector to the plane. From the given line and plane, the direction vector is $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$ and the normal vector is $\vec{n} = 2\hat{i} - \hat{j} + \sqrt{\lambda}\hat{k}$. Calculating the magnitudes gives $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = 3$ and $|\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{5 + \lambda}$.

The dot product is $\vec{b} \cdot \vec{n} = (1)(2) + (2)(-1) + (2)(\sqrt{\lambda}) = 2\sqrt{\lambda}$. Substituting these values into the sine formula, we have $\frac{2\sqrt{\lambda}}{3\sqrt{5 + \lambda}} = \frac{1}{3}$. Multiplying both sides by 3 simplifies the expression to $\frac{2\sqrt{\lambda}}{\sqrt{5 + \lambda}} = 1$. Squaring both sides yields the equation $4\lambda = 5 + \lambda$, which simplifies to $3\lambda = 5$, resulting in $\lambda = \frac{5}{3}$.

Calculating the distance between a line and a parallel plane requires finding the length of the perpendicular segment from any arbitrary point on the line to the plane itself. By identifying the point $(2, -2, 3)$ from the provided line equation and expressing the plane as $x + 5y + z - 5 = 0$, we can apply the standard formula for the distance from a point to a plane. We substitute the point's coordinates into the plane's expression and divide by the magnitude of the normal vector, resulting in $\frac{|2 - 10 + 3 - 5|}{\sqrt{1^2 + 5^2 + 1^2}}$. Evaluating this expression simplifies the numerator to $10$ and the denominator to $\sqrt{27}$, which is equivalent to $3\sqrt{3}$, yielding a final distance of $\frac{10}{3\sqrt{3}}$.

Two lines are perpendicular if the scalar product of their direction vectors equals zero. By expressing the first line $2x = 3y = -z$ in symmetric form as $\frac{x}{3} = \frac{y}{2} = \frac{z}{-6}$, we identify the direction ratios as $(3, 2, -6)$. Similarly, the second line $6x = -y = -4z$ yields the direction ratios $(2, -12, -3)$ by determining the values $x:y:z$ that satisfy the proportionality $x/ (1/6) = y / (-1) = z / (-1/4)$, which simplifies to $2 : -12 : -3$ after multiplying by 12. The scalar product of these two direction vectors is:

$(3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$

Because the scalar product is zero, the lines are perpendicular, indicating an angle of $90^{\circ}$.

In three-dimensional coordinate geometry, any line is defined by the direction angles $\alpha, \beta, \gamma$ it makes with the axes, which must satisfy the fundamental identity:
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$
Given that the line makes an angle $\theta$ with both the x and z axes, this expression simplifies to $2 \cos^2 \theta + \cos^2 \beta = 1$. Rearranging this, we find $2 \cos^2 \theta = 1 - \cos^2 \beta$, and since $1 - \cos^2 \beta = \sin^2 \beta$, we arrive at the relation:
$2 \cos^2 \theta = \sin^2 \beta$
Substituting the provided condition $\sin^2 \beta = 3 \sin^2 \theta$ into this equation yields:
$2 \cos^2 \theta = 3 \sin^2 \theta$
By applying the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$, the equation becomes $2 \cos^2 \theta = 3(1 - \cos^2 \theta)$, which simplifies to $2 \cos^2 \theta = 3 - 3 \cos^2 \theta$. Adding $3 \cos^2 \theta$ to both sides leads to $5 \cos^2 \theta = 3$, resulting in:
$\cos^2 \theta = \frac{3}{5}$

The intersection of two spheres always lies on a flat surface known as the radical plane, which is determined by subtracting the equations of the two spheres from each other. Because the quadratic terms for the coordinates are identical in both sphere equations, subtracting one from the other removes them entirely, leaving behind a linear relationship between $x$, $y$, and $z$ that defines this plane. By arranging the given sphere equations as $x^{2+}y^{2+}z^{2+}7 x-2 y-z-13=0$ and $x^{2+}y^{2+}z^{2-}3 x+3 y+4 z-8=0$, we can perform this subtraction to eliminate the squared terms.

Subtracting the second equation from the first results in $(7 x - (-3 x)) + (-2 y - 3 y) + (-z - 4 z) + (-13 - (-8)) = 0$, which simplifies to $10 x-5 y-5 z-5=0$. Dividing this entire expression by 5 yields the plane equation $2 x-y-z=1$.

Two straight lines in three-dimensional space are co-planar if the scalar triple product of the vector connecting any point on the first line to any point on the second line and the two direction vectors of the lines is equal to zero. Identifying a point and the direction vector for each line by rewriting the equations in symmetric form reveals the components necessary for the condition. For the first line $x=1+s, y=-3-\lambda s, z=1+\lambda s$, a point is $A_1(1, -3, 1)$ with direction vector $V_1(1, -\lambda, \lambda)$, and for the second line $x=t/2, y=1+t, z=2-t$, a point is $A_2(0, 1, 2)$ with direction vector $V_2(1/2, 1, -1)$. The vector connecting the two points is $A_2 - A_1 = (-1, 4, 1)$.

The condition for co-planarity requires the determinant of the matrix formed by these three vectors to be zero:
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ \frac{1}{2} & 1 & -1 \end{vmatrix} = 0$
Applying the column operation of replacing the second column with the sum of the second and third columns yields a matrix with two zeros in the second column:
$\begin{vmatrix} -1 & 5 & 1 \\ 1 & 0 & \lambda \\ \frac{1}{2} & 0 & -1 \end{vmatrix} = 0$
Expanding this determinant along the second column results in $-5(-1 - \frac{\lambda}{2}) = 0$. Solving this equation for $\lambda$ leads to $\lambda = -2$.

The distance between two parallel planes is calculated by determining the perpendicular gap between them after ensuring their normal vectors are aligned. Given the equations $2x+y+2z=8$ and $4x+2y+4z+5=0$, we begin by normalizing the coefficients, dividing the second equation by 2 to obtain $2x+y+2z+\frac{5}{2}=0$, and rewriting the first as $2x+y+2z-8=0$. Once the normal vector components match, the distance formula $d = \frac{|D_1 - D_2|}{\sqrt{A^{2+}B^{2+}C^2}}$ allows us to calculate the value directly. Substituting $D_1 = -8$ and $D_2 = \frac{5}{2}$ alongside the coefficients 2, 1, and 2 into the formula yields $\frac{|\frac{5}{2} - (-8)|}{\sqrt{2^{2+}1^{2+}2^2}}$. This simplifies to the absolute value of 21/2 divided by 3, resulting in a final distance of 7/2.