Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

To find the intersection points on these two given lines, we describe any arbitrary point on the first line $x=y+a=z$ as $(\lambda, \lambda-a, \lambda)$ using a single parameter $\lambda$. We similarly represent an arbitrary point on the second line $x+a=2y=2z$ by setting $2y=2z=\mu$, which allows us to write the point as $(\mu-a, \frac{\mu}{2}, \frac{\mu}{2})$. Because the line segment connecting these two points must align with direction ratios proportional to $2, 1, 2$, the differences between the corresponding coordinates must satisfy the ratio equality $\frac{\lambda-\mu+a}{2} = \frac{\lambda-a-\frac{\mu}{2}}{1} = \frac{\lambda-\frac{\mu}{2}}{2}$.

By equating the first and third expressions, we obtain $\lambda-\mu+a = \lambda-\frac{\mu}{2}$, which simplifies directly to $\frac{\mu}{2}=a$, or $\mu=2a$. Using this value to equate the second and third ratios, we have $2(\lambda-a-\frac{\mu}{2}) = \lambda-\frac{\mu}{2}$, which rearranges to $\lambda-\frac{\mu}{2}=2a$. Substituting $\frac{\mu}{2}=a$ into this expression yields $\lambda-a=2a$, identifying $\lambda=3a$. Replacing these parameter values back into our original coordinate expressions, we find the first point is $(3a, 3a-a, 3a)$ and the second is $(2a-a, \frac{2a}{2}, \frac{2a}{2})$, resulting in $(3a, 2a, 3a)$ and $(a, a, a)$.

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 equals zero. Given the points $(2, 3, 4)$ and $(1, 4, 5)$ from the lines, the connecting vector is $(1, -1, -1)$, which allows us to define the coplanarity condition using the determinant

. Expanding this determinant results in the equation $k^2 + 3k = 0$, which can be simplified to the form $k(k+3) = 0$. Solving this equation reveals that the lines are coplanar when $k$ is $0$ or $-3$.

A cross-section formed by a plane cutting through a sphere creates a circle where the sphere's radius, the plane's perpendicular distance from the sphere's center, and the circle's radius form a right triangle. By rearranging the sphere equation $x^{2+}y^{2+}z^{2+}2x-2y-4z-19=0$ into the standard form $(x+1)^2 + (y-1)^2 + (z-2)^2 = 25$, we identify the center as $(-1, 1, 2)$ and the radius as 5. The perpendicular distance $d$ from this center to the plane $x+2y+2z+7=0$ is given by $d = \frac{|1(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^{2+}2^{2+}2^2}} = \frac{12}{3} = 4$. Using the Pythagorean theorem to relate the sphere radius, the perpendicular distance, and the circle radius, the circle's radius is found by calculating $\sqrt{5^2 - 4^2}$, which results in 3.

When a plane intersects a set of rectangular axes at distances $a, b,$ and $c$ from the origin, its equation is expressed as $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. The perpendicular distance $d$ from the origin $(0,0,0)$ to this plane is determined by the standard geometric formula $d = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$. Rearranging this equation by squaring both sides shows that the square of the reciprocal of the distance is equal to the sum of the squares of the reciprocal intercepts, or $\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. Because the perpendicular distance from a fixed point to a fixed plane is invariant, this value must remain constant regardless of the orientation of the rectangular axes chosen. Therefore, the sum of the inverse squares of the intercepts in the first system must equal the sum of the inverse squares of the intercepts in the second system, resulting in the relationship $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.

The shortest distance between a plane and a sphere is found by determining the perpendicular distance from the sphere's center to the plane and then subtracting the radius of the sphere. Rearranging the sphere equation $x^{2+}y^{2+}z^{2+}4x-2y-6z=155$ into standard form by completing the square yields $(x+2)^{2+}(y-1)^{2+}(z-3)^2=169$, which identifies the center as $(-2, 1, 3)$ and the radius as 13. The perpendicular distance $d$ between the center $(-2, 1, 3)$ and the plane $12x+4y+3z=327$ is given by:
$d = \frac{|12(-2) + 4(1) + 3(3) - 327|}{\sqrt{12^2 + 4^2 + 3^2}}$
Simplifying this expression leads to $d = 26$. Subtracting the radius of 13 from this perpendicular distance of 26 yields a final shortest distance of 13.

Two lines in 3D space are perpendicular if and only if the dot product of their direction vectors equals zero. Expressing the given lines in symmetric form as $\frac{x-b}{a} = \frac{y}{1} = \frac{z-d}{c}$ and $\frac{x-b'}{a'} = \frac{y}{1} = \frac{z-d'}{c'}$ reveals the direction ratios $(a, 1, c)$ and $(a', 1, c')$. Setting the sum of the products of these components to zero results in $a(a') + 1(1) + c(c') = 0$, which simplifies to $aa' + cc' + 1 = 0$.

The line is defined by the set of points where $\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}$ holds true. Because the required plane must contain the entire line, its equation must be satisfied by every point lying on that line. We can identify two such points: the one explicitly given, $(3,2,0)$, and a point derived from the line's own equation, $(4,7,4)$. Substituting the coordinates of these points into the provided options isolates the only consistent choice, where $x-y+z=1$ satisfies both $3-2+0=1$ and $4-7+4=1$.

Any plane defined by the general equation $ax+by+cz=d$ possesses a normal vector with components $(a, b, c)$. Since the plane passes through the point $(1,0,0)$, its equation takes the form $a(x-1) + by + cz = 0$, and the requirement that it also contains $(0,1,0)$ implies $a(0-1) + b(1-0) + c(0) = 0$, which forces $b=a$. The normal vector to the plane is therefore $(a, a, c)$.

Given that the angle between this plane and the reference plane $x+y=3$ is $\frac{\pi}{4}$, we calculate the cosine of this angle using the dot product of the normal vectors divided by the product of their magnitudes. With the reference plane's normal vector being $(1, 1, 0)$, we find $\cos\left(\frac{\pi}{4}\right) = \frac{|(a, a, c) \cdot (1, 1, 0)|}{\sqrt{a^{2+}a^{2+}c^2} \cdot \sqrt{1^{2+}1^2}}$, which simplifies to the equation $\frac{1}{\sqrt{2}} = \frac{2a}{\sqrt{2a^{2+}c^2} \cdot \sqrt{2}}$. Multiplying both sides by $\sqrt{2}$ yields $1 = \frac{2a}{\sqrt{2a^{2+}c^2}}$, and squaring both sides confirms $2a^{2+}c^2 = 4a^2$, or $c^2 = 2a^2$, resulting in $c = \sqrt{2}a$. Consequently, the direction ratios are proportional to $(a, a, \sqrt{2}a)$, which simplifies to $(1, 1, \sqrt{2})$.

The shortest distance between two skew lines is measured along a unique line segment $PQ$ that is mutually perpendicular to both lines. Because this segment must be orthogonal to the direction vectors of both lines, its orientation is parallel to the cross product of these vectors.

To determine the specific coordinates of $P$ and $Q$, we represent arbitrary points on the lines $L_1$ and $L_2$ using the parameters $\lambda$ and $\mu$, resulting in $P = (1+\lambda, 2-\lambda, 3+\lambda)$ and $Q = (4+\mu, 5+\mu, 6-\mu)$. The vector $PQ$ formed by connecting these points is $(3+\mu-\lambda, 3+\mu+\lambda, 3-\mu-\lambda)$. This vector must align with the common normal, found by taking the cross product of the direction vectors $(1, -1, 1)$ and $(1, 1, -1)$, which yields the vector $(0, 2, 2)$. Setting the components of $PQ$ proportional to $(0, 2, 2)$ provides the following system:

$\frac{3+\mu-\lambda}{0} = \frac{3+\mu+\lambda}{2} = \frac{3-\mu-\lambda}{2}$

From the first equality, the numerator $3+\mu-\lambda$ must be $0$, which implies $\lambda-\mu=3$. The second equality simplifies to $3+\mu+\lambda = 3-\mu-\lambda$, leading to $\mu=-\lambda$. Substituting $\mu=-\lambda$ into the first relation gives $2\lambda=3$, so $\lambda = \frac{3}{2}$ and $\mu = -\frac{3}{2}$. Evaluating the coordinates with these parameters gives $P = (\frac{5}{2}, \frac{1}{2}, \frac{9}{2})$ and $Q = (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})$. The midpoint $(\alpha, \beta, \gamma)$ of $PQ$ is the average of these coordinate values, resulting in $(\frac{5}{2}, 2, 6)$. Multiplying the sum of these coordinates by two gives $2(\frac{5}{2} + 2 + 6) = 2(10.5) = 21$.

The cross product of the normal vectors of two planes produces a vector perpendicular to both, which serves as the normal vector for the plane $E$. Given the planes $2x-2y+z=0$ and $x-y+2z=4$ with normal vectors $\vec{n}_1=(2,-2,1)$ and $\vec{n}_2=(1,-1,2)$, the normal vector $\vec{n}_E$ is calculated as:

$\vec{n}_E = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = (-3, -3, 0)$

For algebraic simplicity, we can scale this vector to $(1, 1, 0)$.

Since the plane $E$ passes through $P(1,-1,1)$, we use the point-normal equation $1(x-1) + 1(y+1) + 0(z-1) = 0$, which simplifies to $x+y=0$. The distance from a point $Q(a, a, 2)$ to this plane must equal $3\sqrt{2}$, allowing us to set up the following equation:

$\frac{|a+a|}{\sqrt{1^{2+}1^2}} = 3\sqrt{2}$

Evaluating this expression results in $\frac{|2a|}{\sqrt{2}} = 3\sqrt{2}$, which implies $|2a| = 6$ or $|a|=3$.

With $a$ identified as either $3$ or $-3$, the point $Q$ is $(3, 3, 2)$ or $(-3, -3, 2)$. Calculating the squared distance from $P(1,-1,1)$ to $Q$ yields:

$(PQ)^2 = (a-1)^2 + (a-(-1))^2 + (2-1)^2 = (a-1)^2 + (a+1)^2 + 1$

Substituting $a=3$ gives $2^2 + 4^2 + 1 = 4 + 16 + 1 = 21$. Similarly, substituting $a=-3$ gives $(-4)^2 + (-2)^2 + 1 = 16 + 4 + 1 = 21$. In both instances, the final squared distance is 21.

The shortest distance between two skew lines in three-dimensional space represents the length of the line segment that is perpendicular to both lines. This distance is calculated using the scalar projection of the vector connecting a point on each line onto the common perpendicular vector, given by the formula
$d = \left| \frac{ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) }{ | \vec{b}_1 \times \vec{b}_2 | } \right|$
where $\vec{a}_1$ and $\vec{a}_2$ are the position vectors of points on the lines and $\vec{b}_1$ and $\vec{b}_2$ are their direction vectors.

Converting the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ into their standard symmetric forms yields $\frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}$ and $\frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6}$. Applying the distance formula to these lines leads to $d_1 = 2$.

The second pair of lines consists of $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$. Utilizing the same vector projection method for this second pair results in $d_2 = \frac{12}{\sqrt{3}}$.

The final step involves substituting these values into the expression $\frac{32 \sqrt{3} d_1}{d_2}$. Computing $\frac{32 \sqrt{3} \times 2}{12 / \sqrt{3}}$ simplifies to $\frac{64 \sqrt{3} \times \sqrt{3}}{12}$, which reduces to $\frac{64 \times 3}{12}$, giving a final value of 16.

The plane containing the two lines is defined by a normal vector perpendicular to both direction vectors, $\vec{v_1} = \hat{i}+a\hat{j}-\hat{k}$ and $\vec{v_2} = -\hat{i}+\hat{j}-a\hat{k}$. To find this normal vector, we compute the cross product

, which yields $(1-a^2)\hat{i} + (a+1)\hat{j} + (1+a)\hat{k}$. By factoring out $(1+a)$, we see that this vector is parallel to $(1-a)\hat{i} + \hat{j} + \hat{k}$, which we can use as the normal to the plane.

Because both lines pass through the point $(1, 1, 0)$, the plane equation follows the form $(1-a)(x-1) + 1(y-1) + 1(z-0) = 0$, simplifying to $(1-a)x + y + z = 2-a$. The perpendicular distance from the point $(2, 1, 4)$ to this plane is given by $\left| \frac{(1-a)(2) + 1(1) + 1(4) - (2-a)}{\sqrt{(1-a)^2 + 1^2 + 1^2}} \right| = \sqrt{3}$. This simplifies to the equation $\frac{|5-a|}{\sqrt{a^2 - 2a + 3}} = \sqrt{3}$. Squaring both sides results in $(5-a)^2 = 3(a^2 - 2a + 3)$, which expands to $25 - 10a + a^2 = 3a^2 - 6a + 9$. Rearranging these terms produces the quadratic equation $2a^2 + 4a - 16 = 0$, or $a^2 + 2a - 8 = 0$. Factoring gives $(a+4)(a-2) = 0$, meaning the possible values for $a$ are $2$ and $-4$, making the largest value of $a$ equal to 2.

Any plane passing through the intersection of two given planes $P_1 = 0$ and $P_2 = 0$ belongs to the family of planes defined by the equation $P_1 + \lambda P_2 = 0$. Using the equations $x + y + z - 6 = 0$ and $2x + 3y + 4z + 5 = 0$, the plane $P$ can be expressed as $(x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0$. By substituting the point $(0, 2, -2)$ into this equation to satisfy the condition that $P$ passes through it, we get $(0 + 2 - 2 - 6) + \lambda(0 + 6 - 8 + 5) = 0$, which simplifies to $-6 + 3\lambda = 0$, resulting in $\lambda = 2$.

Substituting $\lambda = 2$ back into the family equation gives the simplified form $5x + 7y + 9z + 4 = 0$ for plane $P$. To find the square of the distance from the point $(12, 12, 18)$ to this plane, we apply the distance formula $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ and square the result:
$d^2 = \frac{(5(12) + 7(12) + 9(18) + 4)^2}{5^2 + 7^2 + 9^2}$
Evaluating the numerator yields $(60 + 84 + 162 + 4)^2 = 310^2$, and the denominator becomes $25 + 49 + 81 = 155$. Consequently, the calculation $d^2 = \frac{310^2}{155}$ simplifies to $310 \times \frac{310}{155}$, which equals $310 \times 2$, giving a final value of 620.

Performing the matrix multiplication transforms the given matrix equation into a system of homogeneous linear equations that define the coordinates of the point $P(\alpha, \beta, \gamma)$. This operation yields the relations $2\alpha + 9\beta + 8\gamma = 0$, $10\alpha + 3\beta + 4\gamma = 0$, and $8\alpha + 8\beta + 8\gamma = 0$.

Dividing the third equation by 8 gives $\alpha + \beta + \gamma = 0$, which implies $\gamma = -\alpha - \beta$. Substituting this expression into the first equation leads to $2\alpha + 9\beta + 8(-\alpha - \beta) = 0$, which simplifies to $2\alpha + 9\beta - 8\alpha - 8\beta = 0$, or $\beta = 6\alpha$. Consequently, we find $\gamma = -\alpha - 6\alpha = -7\alpha$.

Since the point lies on the plane $2x + 4y + 3z = 5$, substituting $\beta = 6\alpha$ and $\gamma = -7\alpha$ into this equation gives $2\alpha + 4(6\alpha) + 3(-7\alpha) = 5$. This resolves to $2\alpha + 24\alpha - 21\alpha = 5$, which simplifies to $5\alpha = 5$, establishing that $\alpha = 1$. With $\alpha$ determined, we find $\beta = 6$ and $\gamma = -7$, leading to the final calculation of $6\alpha + 9\beta + 7\gamma = 6(1) + 9(6) + 7(-7) = 11$.

The shortest distance $d$ between two skew lines defined by $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ represents the projection of the vector connecting points on these lines, $\vec{a}_1 - \vec{a}_2$, onto the line's common normal, which is found via the cross product $\vec{b}_1 \times \vec{b}_2$. This geometric relationship is expressed by the formula
$d = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$

Identify the vectors from the given line equations as $\vec{a}_1 = -\hat{i} + 3\hat{k}$, $\vec{a}_2 = -\hat{j} + 2\hat{k}$, $\vec{b}_1 = \hat{i} - a\hat{j}$, and $\vec{b}_2 = \hat{i} - \hat{j} + \hat{k}$. The difference vector is $\vec{a}_1 - \vec{a}_2 = -\hat{i} + \hat{j} + \hat{k}$, and the common normal vector obtained from the cross product $\vec{b}_1 \times \vec{b}_2$ is $-a\hat{i} - \hat{j} + (a-1)\hat{k}$.

Substituting these into the distance equation with $d = \sqrt{\frac{2}{3}}$ yields $\sqrt{\frac{2}{3}} = \frac{|(-\hat{i} + \hat{j} + \hat{k}) \cdot (-a\hat{i} - \hat{j} + (a-1)\hat{k})|}{\sqrt{(-a)^2 + (-1)^2 + (a-1)^2}}$, which simplifies to $\sqrt{\frac{2}{3}} = \frac{|2a-2|}{\sqrt{2a^{2-}2a+2}}$. Squaring both sides results in $\frac{2}{3} = \frac{4(a-1)^2}{2a^{2-}2a+2}$, leading to the quadratic equation $2a^2 - 5a + 2 = 0$. Factoring this expression gives $(2a-1)(a-2) = 0$, and because $a$ must be an integer, $a = 2$.

The shortest distance between two skew lines lies along a unique line segment that is simultaneously perpendicular to both of them. To identify the specific points $M$ and $N$ that define this segment, we express an arbitrary point on each line using scalar parameters. For the line $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$, any point $M$ is given by $(4\lambda+5, \lambda+4, 3\lambda+5)$, while for the line $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$, any point $N$ is represented as $(12\mu-8, 5\mu-2, 9\mu-11)$.

Because the vector $\overrightarrow{MN}$ must be perpendicular to the direction vectors of both lines, namely $\vec{b}_1 = (4, 1, 3)$ and $\vec{b}_2 = (12, 5, 9)$, it must be parallel to their cross product. Computing this cross product, $\vec{b}_1 \times \vec{b}_2$, results in the vector $(-6, 0, 8)$. By setting the vector $\overrightarrow{MN}$ proportional to this cross product, or equivalently requiring that the dot product of $\overrightarrow{MN}$ with both $\vec{b}_1$ and $\vec{b}_2$ equals zero, we establish a system of linear equations in $\lambda$ and $\mu$. Solving this system yields the values $\lambda = -1$ and $\mu = 1$.

Substituting these parameters back into our expressions for the coordinates, we find the specific points $M(1, 3, 2)$ and $N(4, 3, -2)$. These coordinates correspond directly to the position vectors $\overrightarrow{OM}$ and $\overrightarrow{ON}$ from the origin. The dot product $\overrightarrow{OM} \cdot \overrightarrow{ON}$ is calculated by multiplying corresponding components and summing them: $(1)(4) + (3)(3) + (2)(-2)$, which simplifies to $4 + 9 - 4$, resulting in a final value of 9.

The line where two planes intersect is always perpendicular to both of their normal vectors. The normal vector $\vec{n}_1$ to plane $P_1$ is directly identified as $\langle 2, 1, -3 \rangle$, while the normal vector $\vec{n}_2$ to plane $P_2$ is determined to be $\langle -5, 5, 1 \rangle$ from the provided points.

The direction ratios of the intersection line are found by calculating the cross product of these two normals, defined by the determinant:
$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ -5 & 5 & 1 \end{vmatrix}$
Evaluating this cross product yields $16\hat{i} + 13\hat{j} + 15\hat{k}$. Comparing this result to the given ratios $16, \alpha, \beta$ identifies $\alpha = 13$ and $\beta = 15$, which results in $\alpha + \beta = 28$.

The centroid of a triangle is the arithmetic mean of its vertices. For the triangle with vertices $A(0, \alpha, \alpha)$, $B(\alpha, 0, \alpha)$, and $C(\alpha, \alpha, 0)$, the centroid $G$ is located at $(\frac{2\alpha}{3}, \frac{2\alpha}{3}, \frac{2\alpha}{3})$. A point $D$ moving along the line defined by $y=0$ and $x+z=3$ can be parameterized as $(x, 0, 3-x)$ for any real value $x$. The square of the distance $GD$ is given by:

$GD^2 = \left(x - \frac{2\alpha}{3}\right)^2 + \left(0 - \frac{2\alpha}{3}\right)^2 + \left(3 - x - \frac{2\alpha}{3}\right)^2$

Minimizing this quadratic function with respect to $x$ occurs when $x = 1.5$. Substituting this value into the expression yields $2(1.5 - \frac{2\alpha}{3})^2 + (\frac{2\alpha}{3})^2 = \frac{57}{2}$. Expanding the terms and simplifying results in $\frac{(9-4\alpha)^2}{18} + \frac{8\alpha^2}{18} = \frac{57}{2}$. Multiplying through by 18 removes the denominators and leaves us with $(9-4\alpha)^2 + 8\alpha^2 = 513$.

Continuing the algebraic simplification, we arrive at $81 - 72\alpha + 16\alpha^2 + 8\alpha^2 = 513$, which rearranges to $24\alpha^2 - 72\alpha - 432 = 0$. Dividing the entire equation by 24 yields $\alpha^2 - 3\alpha - 18 = 0$. Factoring this expression gives $(\alpha - 6)(\alpha + 3) = 0$. Because the problem states that $\alpha > 0$, we find that $\alpha = 6$.