Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

By representing the intersection of the two planes $x+2y+z=6$ and $y+2z=4$ through a parameter $\lambda$, we find the line $L$ in parametric form. Expressing the variables as $x = 3\lambda - 2$, $y = 4 - 2\lambda$, and $z = \lambda$, we define a generic point $P(\alpha, \beta, \gamma) = (3\lambda - 2, 4 - 2\lambda, \lambda)$ that lies on this line. Since $P$ is the foot of the perpendicular from the point $Q(3, 2, 1)$ to $L$, the direction vector of $QP$ must be orthogonal to the direction vector of the line, which is $(3, -2, 1)$.

Setting the dot product of these vectors to zero yields the following equation:
$3(3\lambda - 5) - 2(2 - 2\lambda) + 1(\lambda - 1) = 0$
Solving this results in $14\lambda - 20 = 0$, or $\lambda = \frac{10}{7}$. Substituting this value back into the expressions for the coordinates gives $P = (\frac{16}{7}, \frac{8}{7}, \frac{10}{7})$. Finally, calculating $21(\alpha + \beta + \gamma)$ involves multiplying the sum of these coordinates by 21, resulting in $21(\frac{34}{7}) = 102$.

A plane functioning as the perpendicular bisector of a line segment must pass through the segment's midpoint and utilize the vector defined by the segment as its normal. First, locate the midpoint by averaging the endpoints $(4,-3,1)$ and $(2,3,-5)$, resulting in $(3,0,-2)$. Next, determine the direction ratios of the line segment by finding the difference between the coordinates, which yields $(2,-6,6)$; simplifying these by a factor of 2 provides the irreducible normal vector components $(1,-3,3)$.

Using the standard form of a plane equation $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$, substitute the midpoint $(3,0,-2)$ and the normal vector $(1,-3,3)$ to obtain $1(x-3) - 3(y-0) + 3(z+2) = 0$. Simplifying this expression leads to $x - 3y + 3z + 3 = 0$. Comparing this result with $ax + by + cz + d = 0$, the integer coefficients are identified as $a=1, b=-3, c=3,$ and $d=3$. Calculating the sum of the squares of these values gives $1^2 + (-3)^2 + 3^2 + 3^2$, which equals 28.

The orientation of lines defined by equations involving direction cosines $(l, m, n)$ is found by solving for their direction ratios. From the constraint $2l + 2m - n = 0$, replacing $n$ with $2l + 2m$ in the equation $mn + nl + lm = 0$ allows for the calculation of specific direction vectors. Substituting and expanding the second equation yields $m(2l + 2m) + l(2l + 2m) + lm = 0$, which simplifies to:

$2l^2 + 5lm + 2m^2 = 0$

Factoring this quadratic results in $(2l + m)(l + 2m) = 0$, providing two distinct sets of direction ratios. For the first case, $2l + m = 0$ implies $m = -2l$, which results in $n = 2l + 2(-2l) = -2l$, giving the direction vector $(1, -2, -2)$. For the second case, $l + 2m = 0$ implies $l = -2m$, which yields $n = 2(-2m) + 2m = -2m$, resulting in the direction vector $(-2, 1, -2)$. The dot product of these two vectors is $(1)(-2) + (-2)(1) + (-2)(-2) = -2 - 2 + 4 = 0$. Since the dot product is zero, the vectors are orthogonal, establishing that the angle between the lines is $\frac{\pi}{2}$.

The direction of line $l$ is determined by the cross product of the direction vectors of lines $l_1$ and $l_2$. Using the direction vectors $(1, 2, 2)$ for $l_1$ and $(2, 2, 1)$ for $l_2$, the cross product yields the direction vector $(-2, 3, -2)$. Because $l$ passes through the origin, its parametric equations are $x=-2\lambda, y=3\lambda, z=-2\lambda$. Equating these with the parametric form of $l_1$, expressed as $(3+t, -1+2t, 4+2t)$, allows us to solve for the parameters and find the intersection point $P(2, -3, 2)$.

Any point $Q$ on $l_2$ is represented by the coordinates $(3+2s, 3+2s, 2+s)$. Applying the distance formula $PQ = \sqrt{17}$, the squared distance equation is $(3+2s-2)^2 + (3+2s+3)^2 + (2+s-2)^2 = 17$. This simplifies to the quadratic equation $9s^2 + 28s + 20 = 0$, which provides the solutions $s=-2$ and $s=-10/9$. Given the requirement for the point to reside in the first octant, we select $s=-10/9$, which results in the coordinates $(7/9, 7/9, 8/9)$. Summing these coordinates gives $22/9$, and multiplying this sum by 18 leads to the final value of 44.

To determine the length of the projection of a vector onto a plane, one must first identify the normal vector to that plane. Since the plane passes through points $A(1,2,3)$, $B(2,3,1)$, and $C(2,4,2)$, the normal vector $\vec{n}$ is found by calculating the cross product of two vectors lying within the plane, such as $\vec{AB}$ and $\vec{AC}$. These vectors are determined by subtracting coordinates, giving $\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{AC} = (2-1)\hat{i} + (4-2)\hat{j} + (2-3)\hat{k} = \hat{i} + 2\hat{j} - \hat{k}$. The cross product $\vec{AB} \times \vec{AC}$ then yields

.

The length of the projection of vector $\vec{OP}$ onto the plane is effectively the magnitude of the vector's component that lies parallel to the plane. Given $\vec{OP} = 2\hat{i} - \hat{j} + \hat{k}$, one can relate this to the angle $\theta$ between $\vec{OP}$ and the normal vector $\vec{n}$ using the dot product formula $\cos \theta = \frac{\vec{OP} \cdot \vec{n}}{|\vec{OP}| |\vec{n}|}$. The magnitudes of the vectors are $|\vec{OP}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$ and $|\vec{n}| = \sqrt{3^2 + (-1)^2 + 1^2} = \sqrt{11}$, while their dot product is $\vec{OP} \cdot \vec{n} = (2)(3) + (-1)(-1) + (1)(1) = 6 + 1 + 1 = 8$. This results in $\cos \theta = \frac{8}{\sqrt{6}\sqrt{11}} = \frac{8}{\sqrt{66}}$.

Using the geometric relationship where the projection length is $|\vec{OP}| \sin \theta$, one can apply the trigonometric identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$ to finalize the calculation. Substituting the known values leads to $\sqrt{6} \sqrt{1 - (\frac{8}{\sqrt{66}})^2} = \sqrt{6} \sqrt{1 - \frac{64}{66}} = \sqrt{6} \sqrt{\frac{2}{66}}$. Simplifying this expression involves multiplying the square roots, resulting in $\sqrt{\frac{12}{66}}$, which reduces to $\sqrt{\frac{2}{11}}$.

When a line resides within a plane, every point on that line necessarily satisfies the plane's equation, and the line's direction vector must be perpendicular to the plane's normal. Rewriting the line's equation as $\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}$ identifies the point $(1, -4, -2)$ and the direction vector $(-1, 2, 3)$. Applying the condition that this point lies on the plane $lx + my + nz = 0$ results in $l - 4m - 2n = 0$, while the perpendicularity requirement implies $-l + 2m + 3n = 0$. Adding these two equations cancels $l$ and reveals $n = 2m$, which subsequently establishes $l = 8m$, identifying the plane as $8x + y + 2z = 0$. The intersection point of the plane and the segment $AB$ joining $A(-3, -6, 1)$ and $B(2, 4, -3)$ follows the section formula as $(\frac{2k-3}{k+1}, \frac{4k-6}{k+1}, \frac{-3k+1}{k+1})$. Substituting these coordinates into the plane equation produces $8(\frac{2k-3}{k+1}) + (\frac{4k-6}{k+1}) + 2(\frac{-3k+1}{k+1}) = 0$. Simplifying the numerator yields $16k - 24 + 4k - 6 - 6k + 2 = 0$, which reduces to $14k - 28 = 0$, confirming that $k = 2$.

The distance from a point $P(x, y, z)$ to a plane $ax+by+cz+d=0$ is determined by the formula $d = \frac{|ax+by+cz+d|}{\sqrt{a^{2+}b^{2+}c^2}}$. Applying this definition to the three given planes, the squares of the distances from $x+y+z=0$, $lx-nz=0$, and $x-2y+z=0$ result in $\frac{(x+y+z)^2}{3}$, $\frac{(lx-nz)^2}{l^{2+}n^2}$, and $\frac{(x-2y+z)^2}{6}$. According to the problem, the sum of these squared distances is equal to 9, providing the equation:
$\frac{(x+y+z)^2}{3} + \frac{(lx-nz)^2}{l^{2+}n^2} + \frac{(x-2y+z)^2}{6} = 9$

Expanding these terms allows for the grouping of variables $x^2$, $y^2$, $z^2$, $xy$, $yz$, and $xz$. The coefficients for $xy$ and $yz$ calculate to zero, while the remaining terms yield $x^2 (\frac{1}{2} + \frac{l^2}{l^{2+}n^2}) + y^2 + z^2 (\frac{1}{2} + \frac{n^2}{l^{2+}n^2}) + xz (1 - \frac{2ln}{l^{2+}n^2}) = 9$. By comparing this resulting equation to the standard form of the locus $x^{2+}y^{2+}z^2=9$, it is clear that the coefficient of the cross-term $xz$ must vanish. Setting $1 - \frac{2ln}{l^{2+}n^2} = 0$ simplifies to $l^{2+}n^2 = 2ln$, which can be rearranged as $(l-n)^2 = 0$, confirming that $l-n=0$.

Four points are coplanar if the volume of the parallelepiped defined by three vectors originating from one of them is zero, which is satisfied when the scalar triple product of these vectors equals zero. Defining our vectors relative to the point $(1, 5, 35)$, we obtain $PQ = (6, 0, -30)$, $PR = (0, \lambda - 5, -28)$, and $PS = (2\lambda - 1, -4, -33)$. The condition for coplanarity requires the determinant of these components to vanish:
$\begin{vmatrix} 6 & 0 & -30 \\ 0 & \lambda - 5 & -28 \\ 2\lambda - 1 & -4 & -33 \end{vmatrix} = 0$
Expanding this determinant along the first row gives $6[-33(\lambda - 5) - 112] - 30[0 - (\lambda - 5)(2\lambda - 1)] = 0$. Simplifying this expression results in 6[-33\lambda + 165 - 112] + 30\[2\lambda^2 - 11\lambda + 5]$ = 0$. Dividing the entire equation by$6$yields$-33\lambda + 53 + 5(2\lambda^2 - 11\lambda + 5) = 0$, which simplifies to$10\lambda^2 - 88\lambda + 78 = 0$. Reducing this further by dividing by$2$results in the quadratic equation$5\lambda^2 - 44\lambda + 39 = 0$. Since the possible values of$\lambda$are the roots of this equation, their sum is given by the ratio of the negative coefficient of the linear term to the leading coefficient, which is$-\frac{-44}{5}$, resulting in$\frac{44}{5}$.

Represent an arbitrary point on the line using the parameter $\lambda$ as $(2\lambda+1, 3\lambda+2, 6\lambda-1)$. When these expressions for $x$, $y$, and $z$ are substituted into the equation of the plane $2x-y+z=6$, the relationship becomes:
$2(2\lambda+1)-(3\lambda+2)+(6\lambda-1)=6$
Simplifying this equation results in $7\lambda=7$, which establishes that $\lambda=1$. Substituting this value back into the parameterized coordinates yields the intersection point $(3, 5, 5)$.

The square of the distance from this point to $(-1, -1, 2)$ is found by calculating the sum of the squared differences of the respective coordinates, represented as $(3 - (-1))^2 + (5 - (-1))^2 + (5 - 2)^2$. This simplifies to $4^2 + 6^2 + 3^2$, which equals $16 + 36 + 9$, resulting in a final value of 61.

The mirror image of a point across a line is located such that the line acts as the perpendicular bisector of the segment connecting the point and its image. Consequently, the midpoint $M$ of the segment between $P(a, 6, 9)$ and its image $Q(20, b, -a-9)$ must lie on the given line. The coordinates of this midpoint are:
$M = \left( \frac{a+20}{2}, \frac{b+6}{2}, \frac{-a}{2} \right)$

Because $M$ must satisfy the equation of the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9}$, we substitute these coordinates into the line equation to obtain:
$\frac{\frac{a+20}{2}-3}{7} = \frac{\frac{b+6}{2}-2}{5} = \frac{\frac{-a}{2}-1}{-9}$
Simplifying each term yields the equality:
$\frac{a+14}{14} = \frac{b+2}{10} = \frac{a+2}{18}$

To determine the values of $a$ and $b$, we equate the first and third expressions: $\frac{a+14}{14} = \frac{a+2}{18}$. Cross-multiplying gives $18(a+14) = 14(a+2)$, which simplifies to $18a + 252 = 14a + 28$, resulting in $4a = -224$, so $a = -56$. Substituting this value into the relation $\frac{b+2}{10} = \frac{a+2}{18}$ gives $\frac{b+2}{10} = \frac{-54}{18}$, or $\frac{b+2}{10} = -3$. This implies $b+2 = -30$, so $b = -32$. Calculating the final value gives $|a+b| = |-56-32| = 88$.

Finding the mirror image of a point with respect to a plane relies on two key geometric properties: the line segment connecting the point to its image must be perpendicular to the plane, and the midpoint of this segment lies exactly on the plane itself. For the point $(1, 3, 5)$ and the plane $4x - 5y + 2z = 8$, we denote the image as $(\alpha, \beta, \gamma)$. Because the line segment connecting the point to its image is perpendicular to the plane, the ratios of the differences in their coordinates must match the coefficients of the plane's normal vector, allowing us to express the relationship as $\frac{\alpha - 1}{4} = \frac{\beta - 3}{-5} = \frac{\gamma - 5}{2} = \lambda$.

This allows us to describe the image coordinates in terms of a single parameter $\lambda$ as $\alpha = 4\lambda + 1$, $\beta = 3 - 5\lambda$, and $\gamma = 2\lambda + 5$. The midpoint of the segment connecting $(1, 3, 5)$ and $(\alpha, \beta, \gamma)$ is $\left(\frac{\alpha + 1}{2}, \frac{\beta + 3}{2}, \frac{\gamma + 5}{2}\right)$, which simplifies to $(2\lambda + 1, 3 - 2.5\lambda, \lambda + 5)$. Since this midpoint must satisfy the plane equation $4x - 5y + 2z = 8$, we substitute these coordinates into the equation to get $4(2\lambda + 1) - 5(3 - 2.5\lambda) + 2(\lambda + 5) = 8$. Simplifying this expression leads to $8\lambda + 4 - 15 + 12.5\lambda + 2\lambda + 10 = 8$, which reduces to $22.5\lambda = 9$, yielding $\lambda = \frac{2}{5}$. With $\lambda$ determined, the coordinates are $\alpha = \frac{13}{5}$, $\beta = 1$, and $\gamma = \frac{29}{5}$. Calculating the final required value $5(\alpha + \beta + \gamma)$ gives $5\left(\frac{13}{5} + 1 + \frac{29}{5}\right)$, which simplifies to $13 + 5 + 29 = 47$.

Any plane passing through the line of intersection of the planes $3x - 2y + 4z - 7 = 0$ and $x + 5y - 2z + 9 = 0$ follows the family equation $(3x - 2y + 4z - 7) + \lambda(x + 5y - 2z + 9) = 0$. By substituting the point $(1, 4, -3)$ into this expression, we solve for $\lambda$ as $(3(1) - 2(4) + 4(-3) - 7) + \lambda(1 + 5(4) - 2(-3) + 9) = 0$, which simplifies to $-24 + 36\lambda = 0$, giving $\lambda = 2/3$. Substituting $\lambda = 2/3$ back into the original family equation and multiplying by 3 produces $3(3x - 2y + 4z - 7) + 2(x + 5y - 2z + 9) = 0$, simplifying to $11x + 4y + 8z - 3 = 0$. Aligning this with the required form $\alpha x + \beta y + \gamma z + 3 = 0$ requires multiplying the entire equation by -1 to yield $-11x - 4y - 8z + 3 = 0$. Comparing coefficients gives $\alpha = -11$, $\beta = -4$, and $\gamma = -8$, resulting in a sum of -23.

The shortest distance between two skew lines in three-dimensional space represents the length of the segment perpendicular to both lines, which is derived from the projection of the displacement vector connecting any two points on the lines onto the vector defined by the cross product of the direction vectors of the lines. Begin by expressing both line equations in standard symmetric form to extract the passing points and directional components. The first line passes through $(1, 2, 1)$ with direction vector $\vec{v}_1 = 2\hat{i}+\hat{j}+3\hat{k}$, and the second line passes through $(2, \lambda, 3)$ with direction vector $\vec{v}_2 = \hat{i}+2\hat{j}+4\hat{k}$.

Determine the displacement vector connecting these points, given by $\vec{a} = \hat{i} + (\lambda-2)\hat{j} + 2\hat{k}$. Computing the cross product of the direction vectors yields $\vec{v}_1 \times \vec{v}_2 = -2\hat{i} - 5\hat{j} + 3\hat{k}$, which has a magnitude of $\sqrt{38}$. The shortest distance is the magnitude of the scalar triple product divided by the magnitude of this cross product, leading to the equation $\frac{|\vec{a} \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|} = \frac{|14-5\lambda|}{\sqrt{38}}$. Equating this to the given distance of $\frac{1}{\sqrt{38}}$ results in the absolute value equation $|14-5\lambda|=1$. Solving for $\lambda$ provides two solutions, $\lambda = \frac{13}{5}$ and $\lambda = 3$, of which the integral value is $3$.

Two straight lines are co-planar if the scalar triple product of the vector connecting a point on each line and their two direction vectors vanishes. The first line passes through $(k, 2, 3)$ with direction $(1, 2, 3)$, while the second line passes through $(-1, -2, -3)$ with direction $(3, 2, 1)$. The displacement vector between the two points is $(k+1, 4, 6)$, which leads to the determinant condition

. Expanding along the first row results in $(k+1)(2-6) - 4(1-9) + 6(2-6) = 0$, simplifying to $-4(k+1) + 32 - 24 = 0$. This expression reduces to $-4k + 4 = 0$, yielding $k = 1$.

Finding the intersection of a line and a plane requires expressing the line in parametric form to determine the unique coordinates that satisfy the plane equation. For the line passing through Q at (3, -4, -5) and R at (2, -3, 1), the direction vector is found by the difference in coordinates, yielding (-1, 1, 6), which allows any point on the line to be expressed as $(x, y, z) = (3-k, -4+k, -5+6k)$ for a parameter $k$. Substituting these expressions into the plane equation $2x+y+z=7$ leads to $2(3-k) + (-4+k) + (-5+6k) = 7$, which simplifies to $5k - 3 = 7$ and solves to $k=2$. The intersection point T is identified as (1, -2, 7) by substituting $k=2$ into the parametric form. The distance from point P at (3, 4, 4) to T at (1, -2, 7) is determined by the distance formula $\sqrt{(1-3)^2 + (-2-4)^2 + (7-4)^2} = \sqrt{4+36+9} = \sqrt{49} = 7$.

A plane containing one line and running parallel to another is defined by the cross product of the direction vectors of both lines, which serves as the normal vector to the plane. Given the first line passes through $(1, -6, -5)$ with a direction of $(3, 4, 2)$ and the second line has a direction of $(4, -3, 7)$, the equation of the plane can be expressed using the determinant form:
$\begin{vmatrix} x-1 & y+6 & z+5 \\ 3 & 4 & 2 \\ 4 & -3 & 7 \end{vmatrix} = 0$
Substituting the coordinates of the point $(1, -1, \alpha)$ into this equation yields:
$\begin{vmatrix} 1-1 & -1+6 & \alpha+5 \\ 3 & 4 & 2 \\ 4 & -3 & 7 \end{vmatrix} = 0$
This simplifies to:
$\begin{vmatrix} 0 & 5 & \alpha+5 \\ 3 & 4 & 2 \\ 4 & -3 & 7 \end{vmatrix} = 0$
Expanding along the first row results in $0(28+6) - 5(21-8) + (\alpha+5)(-9-16) = 0$. This calculation reduces to $-5(13) - 25(\alpha+5) = 0$, which simplifies to $-65 - 25\alpha - 125 = 0$, or $-25\alpha = 190$. Solving for $5\alpha$ provides $5\alpha = -38$, and taking the absolute value leads to $|5\alpha| = 38$.

A plane is uniquely determined by a point and a line that does not contain it, or by two intersecting lines. In this problem, we identify the mirror image point $B$ of $A(2,3,1)$ with respect to the line $\frac{x+1}{2} = \frac{y-3}{1} = \frac{z+2}{-1}$ to establish a point that lies on the required plane. By representing an arbitrary point $M$ on this line as $(2r-1, r+3, -r-2)$, we calculate the vector $\vec{AM} = (2r-3, r, -r-3)$ and ensure it is orthogonal to the line's direction vector $(2, 1, -1)$. Solving the dot product equation $2(2r-3) + 1(r) - 1(-r-3) = 0$ yields $6r = 3$, giving $r = \frac{1}{2}$, which identifies $M$ as $(0, \frac{7}{2}, -\frac{5}{2})$. Since $M$ is the midpoint of $AB$, $B$ is calculated as $2M - A$, resulting in the coordinates $(-2, 4, -6)$.

With the point $B(-2, 4, -6)$ and the line $L_2$, rewritten as $\frac{x-2}{3} = \frac{y-1}{-2} = \frac{z+1}{1}$, we possess sufficient information to define the plane. This line contains the point $P(2, 1, -1)$ and follows the direction vector $(3, -2, 1)$. We find the normal vector to the plane by computing the cross product of the line's direction vector and the vector $\vec{PB} = (-4, 3, -5)$, which results in $(7, 11, 1)$ based on the determinant

. Substituting these values into the plane equation $7(x-2) + 11(y-1) + 1(z+1) = 0$ simplifies the expression to $7x + 11y + z = 24$. Given $\alpha = 7, \beta = 11$, and $\gamma = 1$, the sum $\alpha + \beta + \gamma$ equals 19.

The direction ratios of the line $L$, given by $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$, are $(2, 1, 1)$. Since the plane $P$ is perpendicular to this line, these values function as the normal vector for the plane, which gives the equation the form $2x+y+z+d=0$. Because $Q$ is the mirror image of the point $(2,3,-1)$ with respect to the line $L$, and the line segment connecting $Q$ and $(2,3,-1)$ is perpendicular to $L$, the point $(2,3,-1)$ also lies on the plane $P$.

Substituting $(2,3,-1)$ into the general plane equation $2x+y+z+d=0$ gives $2(2) + 3 - 1 + d = 0$, which simplifies to $4+3-1+d=0$, leading to $d=-6$. The resulting equation for the plane is $2x+y+z-6=0$. Testing the point $(1,2,2)$ in this equation confirms that $2(1) + 2 + 2 - 6 = 0$, verifying that this point lies on the plane.