Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The perpendicular bisector plane of a line segment represents the locus of all points that are equidistant from the two endpoints. By denoting the given points as $(4, -2, 3)$ and $(2, 4, -1)$, we can identify any point $P(x, y, z)$ on this plane by requiring that its squared distance to both endpoints is equal, or $PA^2 = PB^2$.

Expanding this relationship involves setting up the equation $(x - 4)^2 + (y + 2)^2 + (z - 3)^2 = (x - 2)^2 + (y - 4)^2 + (z + 1)^2$. As we expand these binomials, the quadratic terms $x^2$, $y^2$, and $z^2$ appear on both sides and cancel out, reducing the expression to a linear form. Combining all remaining terms results in $-4x + 12y - 8z = -8$, which simplifies to $x - 3y + 2z = 2$.

We can verify which point lies on this plane by substituting the coordinates from the options into this equation. For the point $(4, 0, -1)$, substituting the values gives $4 - 3(0) + 2(-1)$, which simplifies to $4 - 2 = 2$. Since this result matches the value on the right side of the equation, the point $(4, 0, -1)$ lies on the plane.

The foot of the perpendicular $M$ from a point $P$ to a line is the unique position on the line where the vector $\vec{PM}$ is orthogonal to the line's direction vector. The line passing through the points $(1, -2, 3)$ and $(1, 1, 0)$ has a direction defined by the difference vector $(0, 3, -3)$, which simplifies to $(0, 1, -1)$. We can represent any point $M$ on this line using a parameter $t$ as $(1, -2 + t, 3 - t)$.

Since $M$ is the foot of the perpendicular from $P(4, 2, 3)$, the vector $\vec{PM} = (-3, t - 4, -t)$ must satisfy the condition that its dot product with the line's direction vector is zero:
$0(-3) + 1(t - 4) - 1(-t) = 0$
Solving this equation gives $2t - 4 = 0$, which results in $t = 2$. Substituting $t = 2$ back into our expression for $M$ yields the coordinates $(1, 0, 1)$. Finally, testing these coordinates in the plane equation $2x + y - z = 1$ gives $2(1) + 0 - 1 = 1$, which confirms that the foot of the perpendicular lies on this plane.

The intercept form of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where the coordinate intercepts define the points $(a, 0, 0)$, $(0, b, 0)$, and $(0, 0, c)$. Given that the centroid $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$ of the resulting triangle is $(1, 1, 2)$, it follows that $a=3$, $b=3$, and $c=6$, which produces the plane equation $2x + 2y + z = 6$ upon substitution and simplification. A line perpendicular to this plane shares its normal vector $(2, 2, 1)$ as direction ratios. Utilizing these ratios alongside the centroid coordinates $(1, 1, 2)$ as the passing point, the equation of the line is $\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}$.

The geometric relationship between the fixed plane $23x - 10y - 2z + 48 = 0$ and the plane containing the two lines determines the distance between them. If the planes are not parallel, they will intersect, making the distance zero. Since the problem provides a specific, non-zero distance of $\frac{K}{\sqrt{633}}$, the planes must be parallel, a condition that holds true only when $\lambda = -7$.

Establishing this parallel orientation confirms that the distance between the two surfaces is $\frac{3}{\sqrt{633}}$. Matching this calculated distance with the form provided in the problem shows that the value of $K$ is 3.

Two lines in three-dimensional space are coplanar if and only if the vector joining a specific point on the first line to a point on the second line is coplanar with the direction vectors of both lines. Mathematically, this condition is satisfied when the scalar triple product of these three vectors is equal to zero. For the given lines $L_{1}$ and $L_{2}$, the base points are $(-1, 2, 1)$ and $(-2, -1, -1)$ respectively, and their direction vectors are $(2, -1, 1)$ and $(\alpha, 5-\alpha, 1)$. Defining the vector connecting the base points as $(1, 3, 2)$, the condition for coplanarity is represented by the determinant:

$\begin{vmatrix} 1 & 3 & 2 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{vmatrix} = 0$

Expanding this determinant leads to $1(-1 - (5-\alpha)) - 3(2 - \alpha) + 2(2(5-\alpha) + \alpha) = 0$, which simplifies to $8 + 2\alpha = 0$, yielding $\alpha = -4$. Substituting $\alpha = -4$ into the expression for $L_{2}$ results in the line equation $\frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$. Testing the provided points against this equation, the coordinate $(2, -10, -2)$ satisfies the ratio $\frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$ because $\frac{2+2}{-4} = -1$, $\frac{-10+1}{9} = -1$, and $\frac{-2+1}{1} = -1$, confirming this point lies on the line.

The distance from a point to a plane measured in a specific direction is the length of the segment connecting the starting point to the point where the line with that direction intersects the plane. We can represent any point on the line passing through $(1, -2, 3)$ parallel to the direction $(2, 3, -6)$ as $(2r+1, 3r-2, -6r+3)$ using a scalar parameter $r$. Since the intersection point lies on the plane $x - y + z = 5$, we substitute these coordinates into the plane's equation to obtain $(2r+1) - (3r-2) + (-6r+3) = 5$. Simplifying this leads to $-7r + 6 = 5$, which solves to $r = \frac{1}{7}$.

Substituting this parameter back gives the coordinates of the intersection point as $(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7})$. The distance is then the length of the segment connecting $(1, -2, 3)$ and $(\frac{9}{7}, -\frac{11}{7}, \frac{15}{7})$, which is calculated using the distance formula:

$\sqrt{\left(\frac{9}{7} - 1\right)^2 + \left(-\frac{11}{7} - (-2)\right)^2 + \left(\frac{15}{7} - 3\right)^2}$

$\sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2} = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = 1$

Any point lying on the specified plane must satisfy the linear equation defined by the plane's normal vector and a known point. The plane passes through $(1, 2, 1)$ and $(2, 1, 2)$, containing the displacement vector $\langle 1, -1, 1 \rangle$. Additionally, because the plane is parallel to the line $2x = 3y$ at $z = 1$, it encompasses the direction vector $\langle 3, 2, 0 \rangle$. Computing the cross product of these vectors yields the normal vector $\langle -2, 3, 5 \rangle$. Incorporating these into the point-normal equation $-2(x - 1) + 3(y - 2) + 5(z - 1) = 0$ results in the simplified plane equation $2x - 3y - 5z + 9 = 0$. Validating the coordinates from the point $(-2, 0, 1)$ shows that $2(-2) - 3(0) - 5(1) + 9 = 0$, confirming this point lies on the plane.

To determine the equation of a plane from three points, one must first identify the normal vector, which is perpendicular to the plane. By taking the points $(2, 1, 0)$, $(4, 1, 1)$, and $(5, 0, 1)$, we can form two displacement vectors within the plane: $(4-2, 1-1, 1-0) = (2, 0, 1)$ and $(5-2, 0-1, 1-0) = (3, -1, 1)$. The cross product of these vectors gives the normal vector $\vec{n} = (2, 0, 1) \times (3, -1, 1) = i(0 - (-1)) - j(2 - 3) + k(-2 - 0) = (1, 1, -2)$. Applying the point-normal form with the point $(2, 1, 0)$, the plane equation becomes $1(x-2) + 1(y-1) - 2(z-0) = 0$, which simplifies to $x + y - 2z = 3$.

The image of a point $R(2, 1, 6)$ in this plane lies along a line passing through $R$ in the direction of the normal vector $(1, 1, -2)$. We can represent this line parametrically as $x = k + 2$, $y = k + 1$, and $z = -2k + 6$ for a constant $k$. Substituting these coordinates into the plane equation identifies the intersection point, known as the foot of the perpendicular, which occurs when $(k + 2) + (k + 1) - 2(-2k + 6) = 3$. Simplifying this equation yields $6k - 9 = 3$, or $6k = 12$, resulting in $k = 2$ for the foot of the perpendicular; because the image point is equidistant from the plane as the original point but on the opposite side, we use $k = 4$ to find its coordinates. Applying this to our parametric equations gives $(4 + 2, 4 + 1, -2(4) + 6)$, which evaluates to $(6, 5, -2)$.

When reflecting a point across a plane, two geometric conditions must hold: the midpoint of the line segment joining the original point and its mirror image lies on the plane, and the vector connecting these two points acts as the normal vector to the plane. By averaging the coordinates of $(1, 2, 3)$ and $\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$, the midpoint is found to be $\left(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3}\right)$. The direction of the normal vector can be determined by calculating the difference between these two points, yielding $\left(-\frac{10}{3}, -\frac{10}{3}, -\frac{10}{3}\right)$, which simplifies effectively to the direction $(1, 1, 1)$.

With a normal vector of $(1, 1, 1)$ passing through the midpoint, the plane equation is established by the expression $1(x + \frac{2}{3}) + 1(y - \frac{1}{3}) + 1(z - \frac{4}{3}) = 0$. Expanding this leads to the simple linear relation:

$x + y + z = 1$

Testing the coordinate options against this equation shows that $(1, -1, 1)$ is the choice that satisfies the condition, as $1 + (-1) + 1 = 1$.

A plane passing through the intersection of two given planes can be represented using the family equation $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar parameter. For the planes $x + 4y - z + 7 = 0$ and $3x + y + 5z - 8 = 0$, the resulting expression is $(x + 4y - z + 7) + \lambda(3x + y + 5z - 8) = 0$. Grouping the terms by variables yields $(1+3\lambda)x + (4+\lambda)y + (-1+5\lambda)z + (7-8\lambda) = 0$.

To determine the values of $a$ and $b$ for the specified equation $ax + by + 6z = 15$, we observe that the coefficients of this equation must be proportional to those of our combined plane equation. Setting the ratio of the $z$-coefficients equal to the ratio of the constant terms gives $\frac{-1+5\lambda}{6} = \frac{7-8\lambda}{-15}$. Multiplying both sides by $-30$ simplifies the relationship to $-5(-1 + 5\lambda) = 2(7 - 8\lambda)$, which expands to $5 - 25\lambda = 14 - 16\lambda$, leading to $9\lambda = -9$ or $\lambda = -1$. Substituting $\lambda = -1$ back into the combined expression results in $(1-3)x + (4-1)y + (-1-5)z + (7+8) = 0$, which simplifies to $-2x + 3y - 6z + 15 = 0$, or $2x - 3y + 6z = 15$.

With the plane equation established as $2x - 3y + 6z - 15 = 0$, the perpendicular distance from the point $(3, 2, -1)$ is calculated using the formula $\frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Substituting the known values, we obtain $\frac{|2(3) - 3(2) + 6(-1) - 15|}{\sqrt{2^2 + (-3)^2 + 6^2}}$, which evaluates to $\frac{|6 - 6 - 6 - 15|}{\sqrt{4 + 9 + 36}}$. The calculation simplifies to $\frac{|-21|}{\sqrt{49}}$, which equals $\frac{21}{7}$, resulting in a final distance of $3$.

The foot of the perpendicular from a point to a line creates a geometric relationship where the segment joining the point to its foot is orthogonal to the direction of the line. Since the line passes through both the arbitrary point $(\alpha, 7, 1)$ and the foot $P\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$, the vector $\vec{BP}$ defines the line's direction, and the segment $\vec{AP}$ represents the perpendicular distance from point $A(1, 0, 3)$ to the line. Consequently, the dot product of these two perpendicular vectors must result in zero.

Determining the components of these vectors by subtracting the coordinates of the starting points from the coordinates of the foot $P$ provides $\vec{AP} = \left\langle \frac{2}{3}, \frac{7}{3}, \frac{8}{3} \right\rangle$ and $\vec{BP} = \left\langle \frac{5}{3}-\alpha, -\frac{14}{3}, \frac{14}{3} \right\rangle$. Applying the condition $\vec{AP} \cdot \vec{BP} = 0$ leads to the equation:
$\frac{2}{3}\left(\frac{5}{3}-\alpha\right) + \frac{7}{3}\left(-\frac{14}{3}\right) + \frac{8}{3}\left(\frac{14}{3}\right) = 0$
Expanding this expression gives:
$\frac{10}{9} - \frac{2\alpha}{3} - \frac{98}{9} + \frac{112}{9} = 0$
Combining the constant terms results in $\frac{24}{9} - \frac{2\alpha}{3} = 0$, which simplifies to $\frac{8}{3} = \frac{2\alpha}{3}$. Solving this equation confirms $\alpha = 4$.

The shortest distance between two skew lines is determined by projecting the vector connecting a point on each line onto the common perpendicular, which is defined by the cross product of their direction vectors. The first line, $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$, passes through the point $\vec{a} = (1, -1, 0)$ with a direction vector $\vec{c} = (0, -1, 1)$. The second line, which lies at the intersection of the planes $x+y+z+1=0$ and $2x-y+z+3=0$, can be written in symmetric form as $\frac{x+4/3}{-2/3}=\frac{y-1/3}{-1/3}=\frac{z}{1}$, identifying a point $\vec{b} = (-4/3, 1/3, 0)$ and a direction vector $\vec{d} = (-2/3, -1/3, 1)$.

The distance between these lines is calculated using the formula $\frac{|(\vec{b} - \vec{a}) \cdot (\vec{c} \times \vec{d})|}{|\vec{c} \times \vec{d}|}$. Computing the displacement vector $\vec{b} - \vec{a}$ yields $(-7/3, 4/3, 0)$, while the cross product $\vec{c} \times \vec{d}$ evaluates to $(-2/3, -2/3, -2/3)$. The magnitude of the cross product is $\frac{2}{\sqrt{3}}$, and the absolute value of the dot product between the displacement vector and the cross product is $2/3$, leading to a final shortest distance of $\frac{1}{\sqrt{3}}$.

The normal vector of a plane containing two lines is perpendicular to both, so it is obtained by calculating the cross product of the lines' direction vectors. For direction ratios $1, -2, 2$ and $2, 3, -1$, the normal vector $\vec{n}$ is found via the determinant
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 2 & 3 & -1 \end{vmatrix} = -4\hat{i} + 5\hat{j} + 7\hat{k}$
Using the point $(3, 1, 1)$, the plane equation is constructed as $-4(x - 3) + 5(y - 1) + 7(z - 1) = 0$, which simplifies to $-4x + 12 + 5y - 5 + 7z - 7 = 0$ or $-4x + 5y + 7z = 0$. Since the point $(\alpha, -3, 5)$ lies on the plane, substituting these coordinates yields $-4\alpha + 5(-3) + 7(5) = 0$. Solving this results in $-4\alpha - 15 + 35 = 0$, or $-4\alpha + 20 = 0$, which confirms $\alpha = 5$.

The normal to a plane containing two lines is determined by the cross product of their direction vectors, which uniquely defines the orientation of the plane. For the lines with direction vectors $\vec{v}_1 = (1, 1, 0)$ and $\vec{v}_2 = (0, 1, -1)$, the normal vector $\vec{n}$ is calculated as

. Using the point $(1, 0, 0)$ from the first line, the equation of the plane is $-1(x - 1) + 1(y - 0) + 1(z - 0) = 0$, which simplifies to $x - y - z - 1 = 0$.

To locate the foot of the perpendicular $Q(\alpha, \beta, \gamma)$ from the point $M(1, 0, 1)$ to the plane, we construct a line passing through $M$ in the direction of the normal vector $(1, -1, -1)$. This line is represented by the equations $\frac{\alpha - 1}{1} = \frac{\beta - 0}{-1} = \frac{\gamma - 1}{-1} = k$. Expressing the coordinates of $Q$ in terms of $k$ gives $\alpha = k + 1$, $\beta = -k$, and $\gamma = 1 - k$. Substituting these into the plane equation $(k + 1) - (-k) - (1 - k) - 1 = 0$ yields $3k = 1$, or $k = 1/3$. Consequently, the coordinates are $\alpha = 4/3$, $\beta = -1/3$, and $\gamma = 2/3$. The expression $3(\alpha + \beta + \gamma)$ is then evaluated as $3(\frac{4}{3} - \frac{1}{3} + \frac{2}{3}) = 3(\frac{5}{3})$, which results in 5.

The shortest distance between two skew lines is determined by projecting the vector connecting any two points on the lines onto the vector that is mutually perpendicular to both of them. Identifying the starting points on each line as $(3, 8, 3)$ and $(-3, -7, 6)$ gives a difference vector of $(6, 15, -3)$, while the direction vectors are $(3, -1, 1)$ and $(-3, 2, 4)$. The geometric relationship is encapsulated by the scalar triple product of these three vectors, which is given by:

$\begin{vmatrix} 6 & 15 & -3 \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = 270$

Dividing the absolute value of this product by the magnitude of the cross product of the direction vectors, calculated as $\sqrt{11 \times 29 - 49}$, yields $\frac{270}{\sqrt{270}}$, which simplifies to $3\sqrt{30}$.

Finding the image of a point reflected across a line relies on the principle that the foot of the perpendicular dropped from the original point to the line acts as the midpoint between the original point and its image. We define an arbitrary point on the line $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}$ by equating the ratios to a parameter $\lambda$, resulting in general coordinates $(2\lambda - 1, -2\lambda + 3, -\lambda)$. For this to represent the foot of the perpendicular from $(1, 2, -3)$, the vector connecting these two points must be orthogonal to the line's direction vector of $(2, -2, -1)$, which requires the dot product of these vectors to be zero.

The condition leads to the equation
$2(2\lambda - 2) - 2(-2\lambda + 1) - 1(3 - \lambda) = 0$
which simplifies to $9\lambda = 9$, confirming that $\lambda = 1$. Substituting this value back into our parametric coordinates reveals the foot of the perpendicular to be $(1, 1, -1)$. Finally, using the midpoint formula where the foot $(1, 1, -1)$ is the average of $(1, 2, -3)$ and the image $(a, b, c)$, we find $\frac{a+1}{2} = 1$, $\frac{b+2}{2} = 1$, and $\frac{c-3}{2} = -1$. Solving these equations gives $a = 1$, $b = 0$, and $c = 1$, resulting in a sum $a + b + c = 2$.

The scalar projection of a vector $\vec{PQ}$ onto a vector $\vec{AB}$ represents the length of the shadow cast by $\vec{PQ}$ along the direction of $\vec{AB}$, determined by the formula $|\frac{\vec{PQ} \cdot \vec{AB}}{|\vec{AB}|}|$. Defining the vectors by their endpoints yields $\vec{PQ} = (2-1)\hat{i} + (-4 - (-1))\hat{j} + (11-3)\hat{k} = \hat{i} - 3\hat{j} + 8\hat{k}$ and $\vec{AB} = (3 - (-1))\hat{i} + (-2-2)\hat{j} + (10-3)\hat{k} = 4\hat{i} - 4\hat{j} + 7\hat{k}$. The dot product of these vectors is $(1)(4) + (-3)(-4) + (8)(7) = 4 + 12 + 56 = 72$, while the magnitude of the target vector $\vec{AB}$ is $\sqrt{4^2 + (-4)^2 + 7^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$. Dividing the dot product by this magnitude leads to $72 / 9 = 8.00$.

To find the intersection of a line and a plane, we express the coordinates of an arbitrary point on the line in terms of a single parameter, which must simultaneously satisfy the equation of the plane. By setting the equation of the line equal to a parameter $t$, we can define the coordinates as $(2t + 1, 3t - 1, 4t + 2)$, and substituting these into the plane equation $x + 2y + 3z = 15$ leads to $(2t + 1) + 2(3t - 1) + 3(4t + 2) = 15$.

Simplifying this expression results in $20t + 5 = 15$, which directly yields the value $t = \frac{1}{2}$. Evaluating our coordinate expressions at this specific parameter gives the intersection point $P$ as $\left(2, \frac{1}{2}, 4\right)$. Finally, applying the distance formula for the point $P$ from the origin $(0,0,0)$ gives $\sqrt{2^2 + (\frac{1}{2})^2 + 4^2}$, which simplifies to $\sqrt{4 + 0.25 + 16}$ or $\sqrt{20.25}$, ultimately resulting in $\frac{9}{2}$.

The bisector of the angle between two planes consists of all points equidistant from those planes. To determine the equations for these bisectors, we equate the normalized distance formulas of the given planes, expressed as $\frac{2x - y + 2z - 4}{3} = \pm \frac{x + 2y + 2z - 2}{3}$. This generates two linear equations: $x - 3y - 2 = 0$ and $3x + y + 4z - 6 = 0$. Substituting the point $(2, -4, 1)$ into the second equation yields $3(2) + (-4) + 4(1) - 6$, which simplifies to $6 - 4 + 4 - 6 = 0$, confirming that this specific point lies on the bisecting plane.

The equation of a plane passing through $(0, -1, 0)$ is expressed as $ax + b(y + 1) + cz = 0$, and requiring it to also pass through $(0, 0, 1)$ forces $b + c = 0$, implying $c = -b$. Applying the formula for the angle $\frac{\pi}{4}$ between the plane's normal $\vec{n_1} = (a, b, c)$ and the given normal $\vec{n_2} = (0, 1, -1)$ yields $\cos \frac{\pi}{4} = \frac{|b - c|}{\sqrt{a^2 + b^2 + c^2} \sqrt{2}} = \frac{1}{\sqrt{2}}$, which simplifies to the condition $a^2 = 2b^2$. By choosing $b = \sqrt{2}$ and $c = -\sqrt{2}$, we satisfy the relationship $c = -b$ and find $a = 2$, which identifies the direction ratios of the normal as $(2, \sqrt{2}, -\sqrt{2})$.