Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

A plane containing the $Y$-axis is constrained by the fact that its normal vector must be perpendicular to the direction of that axis, which is defined by the vector $(0, 1, 0)$. Using the general point-normal form for a plane passing through $(1, 2, 3)$, we write the equation as
$a(x-1) + b(y-2) + c(z-3) = 0$
Since the normal vector $(a, b, c)$ is orthogonal to $(0, 1, 0)$, the dot product calculation $a(0) + b(1) + c(0) = 0$ immediately confirms that $b = 0$.

With the equation reduced to $a(x-1) + c(z-3) = 0$, we identify another point on the plane to determine the relationship between the coefficients. Because the $Y$-axis passes through the origin $(0, 0, 0)$, this coordinate must satisfy the equation. Substituting these values leads to $a(0-1) + c(0-3) = 0$, which simplifies to $a = -3c$. Replacing $a$ in the simplified equation with $-3c$ yields $-3c(x-1) + c(z-3) = 0$. After dividing by the non-zero constant $c$, the expression becomes $-3x + 3 + z - 3 = 0$, which resolves to $3x - z = 0$.

Any plane passing through the intersection of the two given planes belongs to a family defined by the equation $(x+2y+3z+1)+\lambda(x-y-z-6)=0$. By grouping the coefficients of $x$, $y$, and $z$, we can rewrite this as $(1+\lambda)x+(2-\lambda)y+(3-\lambda)z+(1-6\lambda)=0$, where the coefficients define the normal vector of the target plane. Because the plane is perpendicular to $-2x+y+z+8=0$, the dot product of their respective normal vectors must equal zero, leading to the condition $-2(1+\lambda)+1(2-\lambda)+1(3-\lambda)=0$. Simplifying this expression results in $-2-2\lambda+2-\lambda+3-\lambda=0$, which yields $3-4\lambda=0$ and fixes the value of $\lambda$ at $\frac{3}{4}$. Substituting this value back into the general family equation results in the final plane equation $7x+5y+9z=14$. Substituting the coordinates of the options into this equation reveals that the point $(0,1,1)$ satisfies the requirement since $0+5+9$ equals $14$.

The condition of perpendicularity for two lines is that the dot product of their direction vectors must be zero. Given the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$, we can represent an arbitrary point on this line by setting it equal to a parameter $\lambda$, yielding coordinates $B(2\lambda+1, 3\lambda-1, -2\lambda+1)$.

Since the target line passes through $A(0, 1, 2)$ and $B$, its direction vector $\vec{AB}$ is $(2\lambda+1, 3\lambda-2, -2\lambda-1)$. Applying the perpendicular condition with the direction vector $(2, 3, -2)$ gives:
$2(2\lambda+1) + 3(3\lambda-2) - 2(-2\lambda-1) = 0$
This simplifies to $17\lambda - 2 = 0$, which solves to $\lambda = \frac{2}{17}$.

Substituting $\lambda = \frac{2}{17}$ into the vector components produces $(21/17, -28/17, -21/17)$, which simplifies to the proportional direction ratios $(-3, 4, 3)$. With the point $(0, 1, 2)$ and these ratios, the line equation is expressed as:
$\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$

Any plane passing through the line of intersection of two given planes can be represented by the family of planes equation $P_1 + \lambda P_2 = 0$, where $P_1$ and $P_2$ are the expressions for the two initial planes and $\lambda$ is a scalar parameter. Applying this to the provided equations, we write the expression as $(2 x-7 y+4 z-3) + \lambda(3 x-5 y+4 z+11) = 0$. Because the required plane must pass through the specific point $(-2,1,3)$, we substitute these coordinates for $x$, $y$, and $z$ into the equation to determine the value of $\lambda$, which yields $(-4-7+12-3) + \lambda(-6-5+12+11) = 0$. Simplifying the arithmetic, we find $-2 + 12\lambda = 0$, leading to $\lambda = \frac{1}{6}$.

Substituting this value back into the family equation gives $(2 x-7 y+4 z-3) + \frac{1}{6}(3 x-5 y+4 z+11) = 0$. To simplify, we multiply the entire equation by $6$ to remove the fraction, resulting in $6(2 x-7 y+4 z-3) + (3 x-5 y+4 z+11) = 0$, which expands to $12 x-42 y+24 z-18 + 3 x-5 y+4 z+11 = 0$. Combining the like terms produces the final equation of the plane, $15 x-47 y+28 z-7 = 0$. Comparing this result with the given form $a x+b y+c z-7 = 0$, we identify the coefficients as $a = 15$, $b = -47$, and $c = 28$. Evaluating the final expression $2 a+b+c-7$ involves calculating $2(15) - 47 + 28 - 7$, which simplifies to $30 - 47 + 28 - 7$ and equals $4$.

To determine the equation of a plane containing a line and a given point, identify the plane's normal vector by taking the cross product of the line's direction vector and a vector connecting any point on the line to the external point. Once the standard equation of the plane is established, the distance from the origin is calculated using the perpendicular distance formula for a plane.

Using the line $\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$, we identify a point $B(2, 3, -2)$ and direction vector $\vec{l} = -3\hat{i} + 2\hat{j} + \hat{k}$. Connecting $B$ to the point $A(3, 7, -7)$ yields the vector $\vec{BA} = (3-2)\hat{i} + (7-3)\hat{j} + (-7 - (-2))\hat{k} = \hat{i} + 4\hat{j} - 5\hat{k}$. Computing the cross product $\vec{l} \times \vec{BA}$ results in the normal vector $\vec{n} = -14\hat{i} - 14\hat{j} - 14\hat{k}$, which scales to $\hat{i} + \hat{j} + \hat{k}$. Substituting these components into the plane equation $(x-2) + (y-3) + (z+2) = 0$ simplifies to $x + y + z - 3 = 0$. The perpendicular distance $d$ from the origin $(0, 0, 0)$ is $\frac{|-3|}{\sqrt{1^2 + 1^2 + 1^2}} = \sqrt{3}$, which implies $d^2 = 3$.

The shortest distance between two skew lines in space is derived by calculating the scalar projection of the vector connecting arbitrary points on the lines onto the common perpendicular vector, which is defined by the cross product of the two direction vectors of the lines. We first convert the equations of the given lines into symmetric form to easily identify their points and direction vectors. The first line $x-\lambda = 2y-1 = -2z$ rearranges to $\frac{x-\lambda}{2} = \frac{y-1/2}{1} = \frac{z}{-1}$, passing through point $P_1 = (\lambda, 1/2, 0)$ with direction vector $\vec{b}_1 = \langle 2, 1, -1 \rangle$. The second line $x = y+2\lambda = z-\lambda$ becomes $\frac{x}{1} = \frac{y-(-2\lambda)}{1} = \frac{z-\lambda}{1}$, which passes through point $P_2 = (0, -2\lambda, \lambda)$ with direction vector $\vec{b}_2 = \langle 1, 1, 1 \rangle$.

The common perpendicular vector is the cross product $\vec{b}_1 \times \vec{b}_2 = \langle 2, -3, 1 \rangle$, which has a magnitude of $\sqrt{14}$. Substituting these components into the distance formula $\frac{|(P_2 - P_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ using the displacement vector $P_2 - P_1 = \langle -\lambda, -2\lambda - 1/2, \lambda \rangle$ results in the expression $\frac{|5\lambda + 3/2|}{\sqrt{14}}$. Equating this to the given shortest distance $\frac{\sqrt{7}}{2\sqrt{2}} = \frac{7}{2\sqrt{14}}$ yields the equation $|5\lambda + 3/2| = 7/2$, or equivalently $|10\lambda + 3| = 7$. This absolute value equation gives two possible values for $\lambda$, specifically $\lambda = 0.4$ and $\lambda = -1$. Given that $\lambda$ must be an integer, we conclude that $\lambda = -1$, and therefore $|\lambda| = 1$.

The mirror image of a point across a plane relies on two fundamental geometric properties: the line segment connecting the point and its image must be perpendicular to the plane, and the midpoint of this segment must lie directly on the plane. For the point $P(1, 3, a)$ and its image $Q(-3, 5, 2)$, the midpoint $R$ is calculated as $(\frac{1-3}{2}, \frac{3+5}{2}, \frac{a+2}{2})$, which simplifies to $(-1, 4, \frac{a+2}{2})$. Because this midpoint sits on the plane defined by $2x - y + z = b$, we substitute these coordinates into the plane equation to get $2(-1) - 4 + \frac{a+2}{2} = b$, which rearranges to the relationship $a = 2b + 10$.

Simultaneously, the vector connecting the point and its image, $\vec{PQ} = (1 - (-3), 3 - 5, a - 2) = (4, -2, a - 2)$, must be parallel to the plane's normal vector, $(2, -1, 1)$. This proportionality implies that $\frac{4}{2} = \frac{-2}{-1} = \frac{a-2}{1}$, which simplifies to $\frac{a-2}{1} = 2$, identifying that $a = 4$. Substituting this value back into the earlier linear relationship gives $4 = 2b + 10$, resulting in $2b = -6$ or $b = -3$. Finally, the value of $|a + b|$ is calculated as $|4 + (-3)|$, which equals $1$.

The family of planes passing through the intersection of two planes, defined by $P_1 = 0$ and $P_2 = 0$, is expressed by the equation $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar. For the given planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (\hat{i} - 2\hat{j}) = -2$, we first express them as $r \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 = 0$ and $r \cdot (\hat{i} - 2\hat{j}) + 2 = 0$. Combining these into the general equation for the required plane gives $[r \cdot (\hat{i} + \hat{j} + \hat{k}) - 1] + \lambda [r \cdot (\hat{i} - 2\hat{j}) + 2] = 0$.

Since the plane passes through the point $(1, 0, 2)$, we substitute $r = \hat{i} + 2\hat{k}$ into this equation to solve for $\lambda$. This yields $[(\hat{i} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) - 1] + \lambda [(\hat{i} + 2\hat{k}) \cdot (\hat{i} - 2\hat{j}) + 2] = 0$. Calculating the dot products, we get $(1 + 0 + 2 - 1) + \lambda (1 + 0 + 0 + 2) = 0$, which simplifies to $2 + 3\lambda = 0$, or $\lambda = -\frac{2}{3}$.

Substituting $\lambda = -\frac{2}{3}$ back into the family equation, we obtain $[r \cdot (\hat{i} + \hat{j} + \hat{k}) - 1] - \frac{2}{3} [r \cdot (\hat{i} - 2\hat{j}) + 2] = 0$. To clear the fraction, we multiply the entire equation by 3, resulting in $3[r \cdot (\hat{i} + \hat{j} + \hat{k}) - 1] - 2[r \cdot (\hat{i} - 2\hat{j}) + 2] = 0$. Expanding the terms gives $r \cdot (3\hat{i} + 3\hat{j} + 3\hat{k} - 2\hat{i} + 4\hat{j}) = 3 + 4$, which simplifies to the final vector equation:

$r \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$

The shortest distance from a point to a plane is the length of the perpendicular segment $PQ$, which requires first establishing the equation of the plane containing the given lines. The plane's normal vector is the cross product of the direction vectors of the lines, $\langle 6, 7, 8 \rangle$ and $\langle 3, 5, 7 \rangle$, which results in $\langle 9, -18, 9 \rangle$ and simplifies to the proportional vector $\langle 1, -2, 1 \rangle$, yielding the plane equation $x-2y+z=0$. Substituting the coordinates of $P(7, -2, 13)$ into the distance formula $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^{2+}B^{2+}C^2}}$ gives $PQ = \frac{|1(7)-2(-2)+1(13)|}{\sqrt{1^{2+}(-2)^{2+}1^2}} = \frac{24}{\sqrt{6}}$. Squaring this result provides $PQ^2 = 96$.

The reflection of a point across a plane lies on a line perpendicular to the plane, following the relationship $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2\frac{ax_1+by_1+cz_1+d}{a^{2+}b^{2+}c^2}$. Applying this to the point $Q(1, 3, 4)$ and the plane $2x - y + z + 3 = 0$, we find the ratios $\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = -2\frac{2(1)-3+4+3}{2^{2+}(-1)^{2+}1^2}$. Simplifying the right side of this equation to $-2$ identifies the coordinates of the mirror image $S$ as $(-3, 5, 2)$.

Since $R(3, 5, \gamma)$ lies on the plane, substituting these values into the plane equation $2x - y + z + 3 = 0$ gives $2(3) - 5 + \gamma + 3 = 0$. Solving this simplifies to $4 + \gamma = 0$, confirming $\gamma = -4$, so $R$ is positioned at $(3, 5, -4)$. To find the square of the length of segment $SR$, we use the squared distance formula $(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$ with points $S(-3, 5, 2)$ and $R(3, 5, -4)$. This calculation $(3 - (-3))^2 + (5 - 5)^2 + (-4 - 2)^2$ simplifies to $6^2 + 0^2 + (-6)^2$, which totals 72.

The equation of the plane containing the points $(1,0,1)$, $(1,-2,1)$, and $(0,1,-2)$ is determined by finding the normal vector through the cross product of two vectors lying within the plane. Subtracting the points yields vectors $(0, -2, 0)$ and $(-1, 1, -3)$, whose cross product is $6\hat{i}-2\hat{k}$. Simplifying this normal vector to $3\hat{i}-\hat{k}$, we construct the plane equation as $3(x-1) - 1(z-1) = 0$, which simplifies to $3x-z-2=0$.

Since the vector $\vec{a} = \alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ is parallel to this plane, its dot product with the normal vector $(3,0,-1)$ must equal zero, establishing the relation $3\alpha - \gamma = 0$, or $\gamma = 3\alpha$. The condition that $\vec{a}$ is perpendicular to $\hat{i}+2\hat{j}+3\hat{k}$ implies $\alpha + 2\beta + 3\gamma = 0$. Substituting $\gamma=3\alpha$ into this expression gives $\alpha + 2\beta + 9\alpha = 0$, which simplifies to $\beta = -5\alpha$.

Utilizing the third condition, $\vec{a} \cdot (\hat{i}+\hat{j}+2\hat{k}) = 2$, we arrive at the equation $\alpha + \beta + 2\gamma = 2$. By substituting the relationships $\beta = -5\alpha$ and $\gamma = 3\alpha$ into this equation, we get $\alpha - 5\alpha + 2(3\alpha) = 2$, which reduces to $2\alpha = 2$, revealing that $\alpha = 1$. With $\alpha$ determined, we find $\beta = -5$ and $\gamma = 3$. Evaluating the required expression $(\alpha-\beta+\gamma)^2$ with these values results in $(1 - (-5) + 3)^2 = 9^2$, which equals 81.

The geometry of this problem relies on finding the coordinates of points $N$ and $Q$ by applying vector orthogonality conditions. Since point $N$ lies on the line $L$ given by $\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$, we denote its coordinates as $(k, 0, -k)$. For $N$ to be the foot of the perpendicular from $P(1, 2, -1)$, the vector $\vec{PN} = (k-1, -2, -k+1)$ must be perpendicular to the line's direction vector $\hat{i} - \hat{k}$. Solving the dot product $(k-1, -2, -k+1) \cdot (1, 0, -1) = 0$ yields $k=1$, identifying $N$ as $(1, 0, -1)$ and resulting in a magnitude of $2$ for the vector segment $\vec{PN}$.

Point $Q$ also lies on line $L$, so its coordinates are $(\mu, 0, -\mu)$. Given that the line $PQ$ is parallel to the plane $x+y+2z=0$, the vector $\vec{PQ} = (\mu-1, -2, -\mu+1)$ must be orthogonal to the plane's normal vector $\hat{i} + \hat{j} + 2\hat{k}$. Setting the dot product $(\mu-1, -2, -\mu+1) \cdot (1, 1, 2) = 0$ results in $\mu - 1 - 2 - 2\mu + 2 = 0$, giving $\mu = -1$. This identifies point $Q$ as $(-1, 0, 1)$, which corresponds to the vector $\vec{PQ} = (-2, -2, 2)$.

With these vectors established, the acute angle $\alpha$ between the lines $PN$ and $PQ$ is determined using the cosine formula $\cos \alpha = \frac{|\vec{PN} \cdot \vec{PQ}|}{|\vec{PN}| |\vec{PQ}|}$. Using $\vec{PN} = (0, -2, 0)$ and $\vec{PQ} = (-2, -2, 2)$, the dot product magnitude is $|(0)(-2) + (-2)(-2) + (0)(2)| = 4$. Given the magnitudes $|\vec{PN}| = 2$ and $|\vec{PQ}| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$, the value of $\cos \alpha$ is $\frac{4}{2(2\sqrt{3})}$, which simplifies to $\frac{1}{\sqrt{3}}$.

Finding the distance between a point and the intersection of two planes begins by determining the direction vector of the line formed by that intersection. Because this line must be perpendicular to the normals of both planes, taking the cross product of $\vec{n_1} = (2, 3, 2)$ and $\vec{n_2} = (1, -2, 1)$ yields the vector $(7, 0, -7)$, which simplifies to the direction vector $(1, 0, -1)$. This allows any point on the line to be expressed in parametric form as $Q = (\lambda, 0, -\lambda)$.

The shortest distance from the point $P = (-1, 2, -2)$ to the line occurs when the vector $\vec{PQ} = (\lambda+1, -2, 2-\lambda)$ is orthogonal to the line's direction vector $(1, 0, -1)$. By setting their dot product equal to zero, we solve $(\lambda+1)(1) + (-2)(0) + (2-\lambda)(-1) = 0$ to find $\lambda = \frac{1}{2}$, which establishes the coordinates of $Q$ as $(\frac{1}{2}, 0, -\frac{1}{2})$.

Calculating the magnitude of $\vec{PQ}$ using the coordinates $P$ and $Q$ results in $\vec{PQ} = (\frac{1}{2} - (-1), 0 - 2, -\frac{1}{2} - (-2))$, which simplifies to the vector $(\frac{3}{2}, -2, \frac{3}{2})$. The distance is then determined by $\sqrt{(\frac{3}{2})^2 + (-2)^2 + (\frac{3}{2})^2}$, which equals $\sqrt{\frac{9}{4} + 4 + \frac{9}{4}}$, ultimately simplifying to $\frac{\sqrt{34}}{2}$.

The plane passing through $(42,0,0)$, $(0,42,0)$, and $(0,0,42)$ follows the linear equation $\frac{x}{42} + \frac{y}{42} + \frac{z}{42} = 1$, which simplifies to $x+y+z=42$. By introducing substitution variables $a=x-11$, $b=y-19$, and $c=z-12$, the constraint $x+y+z=42$ transforms into $a+b+c=0$. This observation is critical because the algebraic identity $a^{3+}b^{3+}c^3=3abc$ holds true whenever the sum of three terms is zero.

The expression can be written in terms of these new variables as $3 + \frac{a}{b^2c^2} + \frac{b}{a^2c^2} + \frac{c}{a^2b^2} - \frac{42}{14abc}$. Combining the three fractional terms using the common denominator $a^2b^2c^2$ results in the numerator $a^{3+}b^{3+}c^3$. Since $a+b+c=0$, this numerator is equal to $3abc$, which transforms the combined fraction into $\frac{3abc}{a^2b^2c^2}$, simplifying further to $\frac{3}{abc}$. Simultaneously, the final term of the original expression, $\frac{42}{14abc}$, reduces to $\frac{3}{abc}$. Subtracting this term from the combined fraction nullifies both components, leaving 3 as the final result.

Since the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ is entirely contained within the plane $x+3y-2z+\beta=0$, any point on the line must also satisfy the equation of the plane. By inspecting the symmetric form of the line, we can see that it passes through the point $(2, 2, -2)$. Substituting these coordinates into the plane equation yields $2 + 3(2) - 2(-2) + \beta = 0$, which simplifies to $12 + \beta = 0$, meaning $\beta = -12$.

Additionally, because the line lies within the plane, its direction vector $(\alpha, -5, 2)$ must be perpendicular to the normal vector of the plane, which is $(1, 3, -2)$. For two vectors to be perpendicular, their dot product must equal zero, leading to the equation $\alpha(1) + (-5)(3) + 2(-2) = 0$. Solving this provides $\alpha - 15 - 4 = 0$, which confirms $\alpha = 19$. Combining these results gives the final value $\alpha + \beta = 19 + (-12) = 7$.

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 equation $P_1 + \lambda P_2 = 0$. Given the planes $x + y + z - 1 = 0$ and $2x + 3y - z + 4 = 0$, their intersection is represented by $(x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0$. We can rewrite this expression by grouping the variables as $x(1 + 2\lambda) + y(1 + 3\lambda) + z(1 - \lambda) + (4\lambda - 1) = 0$.

Since the resulting plane is parallel to the X-axis, its normal vector must be perpendicular to the X-axis, which means the coefficient of $x$ in the equation must be zero. Setting $1 + 2\lambda = 0$ allows us to determine the constant $\lambda = -1/2$. Substituting this value back into the combined equation, we get $y(1 - 3/2) + z(1 + 1/2) + (4(-1/2) - 1) = 0$, which simplifies to $-(1/2)y + (3/2)z - 3 = 0$. Multiplying the entire equation by $-2$ results in $y - 3z + 6 = 0$, which is equivalent to the vector form $r \cdot (\hat{j} - 3\hat{k}) + 6 = 0$.

The shortest distance from the point $(2, -1, 6)$ to a line is determined by finding the perpendicular distance from the point to the line. First, rewrite the system of planes $3y - 2z - 1 = 0$ and $3x - z + 4 = 0$ in symmetric form to identify a point $Q$ and the line's direction vector $\vec{d}$. Rearranging the equations gives $\frac{x + 4/3}{1/3} = \frac{y - 1/3}{2/3} = \frac{z - 0}{1}$, which reveals a point $Q = (-4/3, 1/3, 0)$ and a direction vector $\vec{d} = (1/3, 2/3, 1)$.

Define vector $\vec{QP}$ as $(2 - (-4/3), -1 - 1/3, 6 - 0) = (10/3, -4/3, 6)$. The length of the projection of $\vec{QP}$ onto the line, $PR$, is calculated using the dot product formula $PR = \frac{|\vec{QP} \cdot \vec{d}|}{|\vec{d}|}$. Since $|\vec{d}| = \sqrt{(1/3)^2 + (2/3)^2 + 1^2} = \sqrt{14}/3$ and $\vec{QP} \cdot \vec{d} = (10/3)(1/3) + (-4/3)(2/3) + 6(1) = 56/9$, we obtain $PR = \frac{56/9}{\sqrt{14}/3} = \frac{4\sqrt{14}}{3}$. Applying the Pythagorean theorem where the perpendicular distance $QR = \sqrt{|\vec{QP}|^2 - PR^2}$, we find $|\vec{QP}|^2 = (10/3)^2 + (-4/3)^2 + 6^2 = 100/9 + 16/9 + 324/9 = 440/9$. Substituting the values yields $QR = \sqrt{440/9 - 224/9} = \sqrt{216/9} = \sqrt{24}$, which simplifies to $2 \sqrt{6}$.