Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The reflection $Q$ of point $P(1, 2, 3)$ in the plane $2x - y + z = 9$ lies along the line passing through $P$ that is perpendicular to the plane. Given the normal vector $\vec{n} = (2, -1, 1)$, the coordinates of any point on this line can be expressed as $(1+2k, 2-k, 3+k)$. Substituting these coordinates into the plane equation $2x - y + z = 9$ provides $2(1+2k) - (2-k) + (3+k) = 9$, which simplifies to $6k + 3 = 9$, resulting in $k=1$. Because the image point $Q$ is located at twice the distance of the intersection point from $P$ along the normal, we set $k=2$ to obtain the coordinates $Q(5, 0, 5)$.

With the position of $Q$ established, the area of triangle $PQR$ is determined by the cross product of the vectors $\vec{PQ}$ and $\vec{PR}$. We calculate the vectors as $\vec{PQ} = (5-1, 0-2, 5-3) = (4, -2, 2)$ and $\vec{PR} = (6-1, 10-2, 7-3) = (5, 8, 4)$. The cross product $\vec{PQ} \times \vec{PR}$ yields the vector $(-24, -6, 42)$. The area of the triangle is defined as $\frac{1}{2} |\vec{PQ} \times \vec{PR}|$, which corresponds to the magnitude of the vector $(-12, -3, 21)$. Calculating this magnitude gives $\sqrt{(-12)^2 + (-3)^2 + 21^2} = \sqrt{144 + 9 + 441} = \sqrt{594}$. Thus, the square of the area of triangle $PQR$ is 594.

The plane $P$ is determined by two vectors lying within it: the direction vector of the line, $\langle 1, -3, 7 \rangle$, and a vector connecting the line's known point $(1, 2, -5)$ to the external point $(2, 4, -3)$, which is $\langle 1, 2, 2 \rangle$. The normal to the plane is the cross product of these two vectors, yielding $\langle -20, 5, 5 \rangle$, which simplifies to $\langle 4, -1, -1 \rangle$. Using the point $(1, 2, -5)$ and this normal vector, the equation of the plane $P$ is $4(x-1) - 1(y-2) - 1(z+5) = 0$, or simply $4x - y - z = 7$.

Calculating the image $(\alpha, \beta, \gamma)$ of the point $(-1, 3, 4)$ in this plane requires the reflection formula $\frac{\alpha - (-1)}{4} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{-1} = \frac{-2(4(-1) - 3 - 4 - 7)}{4^2 + (-1)^2 + (-1)^2}$. Evaluating the constant on the right side produces $\frac{-2(-18)}{18}$, which results in a ratio of $2$. Equating this ratio to each coordinate expression, we solve $\alpha = 4(2) - 1 = 7$, $\beta = -1(2) + 3 = 1$, and $\gamma = -1(2) + 4 = 2$. Summing these values gives $\alpha + \beta + \gamma = 7 + 1 + 2 = 10$.

The orientation of a plane is determined by two vectors that lie within it. The first vector is defined by the two points $(0,-1,2)$ and $(-1,2,1)$, which is $\langle -1, 3, -1 \rangle$. The second vector is derived from the line connecting $(5,1,-7)$ and $(1,-1,-1)$, yielding the direction $\langle 4, 2, -6 \rangle$. Since both vectors are parallel to the plane, their cross product generates the plane's normal vector.

Calculating the cross product gives

, which simplifies by a factor of $-2$ to the normal vector $\langle 8, 5, 7 \rangle$. Using this normal vector with the point $(0,-1,2)$, the equation of the plane is $8(x-0) + 5(y+1) + 7(z-2) = 0$, which simplifies to $8x + 5y + 7z = 9$. Testing the coordinates $(-2,5,0)$ in this equation results in $8(-2) + 5(5) + 7(0) = 9$, which verifies that this specific point lies on the plane.

Any point on $L_1$ can be expressed as $P(2\lambda+1, \lambda+3, 2\lambda+2)$ and on $L_2$ as $Q(\mu+2, 2\mu+2, 3\mu+3)$ using parameters $\lambda$ and $\mu$. Since the line segment $PQ$ has direction ratios $1, -1, -2$, the vector $\vec{PQ} = (\mu - 2\lambda + 1, 2\mu - \lambda - 1, 3\mu - 2\lambda + 1)$ must be proportional to this direction vector:
$\frac{\mu - 2\lambda + 1}{1} = \frac{2\mu - \lambda - 1}{-1} = \frac{3\mu - 2\lambda + 1}{-2}$
Solving this system yields $\lambda = 3$ and $\mu = 3$, identifying the points as $P(7, 6, 8)$ and $Q(5, 8, 12)$. Consequently, the distance between these points is $PQ = \sqrt{(5-7)^2 + (8-6)^2 + (12-8)^2} = \sqrt{4 + 4 + 16} = 2\sqrt{6}$.

The equation of a plane passing through the line of intersection of two given planes can be determined using the family of planes formula, $P_1 + \lambda P_2 = 0$, where $\lambda$ is a parameter. By setting the two given planes as $2x - y + z - 3 = 0$ and $4x - 3y + 5z + 9 = 0$, the family is represented as $(2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0$. Grouping the variables allows us to express this equation in the standard form $(2 + 4\lambda)x - (1 + 3\lambda)y + (1 + 5\lambda)z = 3 - 9\lambda$. The normal vector to this plane is given by $\vec{n} = (2 + 4\lambda, -(1 + 3\lambda), 1 + 5\lambda)$.

Since the plane is parallel to the line defined by direction ratios $(-2, 4, 5)$, the normal vector of the plane must be perpendicular to the direction vector of the line. This requires the dot product of these two vectors to be zero, leading to $-2(2 + 4\lambda) + 4(-(1 + 3\lambda)) + 5(1 + 5\lambda) = 0$. Expanding this yields $-4 - 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$, which simplifies to $5\lambda - 3 = 0$, revealing that $\lambda = 3/5$. Substituting this value of $\lambda$ back into the grouped equation results in $\frac{22}{5}x - \frac{14}{5}y + 4z = 3 - \frac{27}{5}$, which simplifies to $\frac{22}{5}x - \frac{14}{5}y + 4z = -\frac{12}{5}$. Multiplying the entire equation by $5/2$ provides $11x - 7y + 10z = -6$, or $11x - 7y + 10z + 6 = 0$. Comparing this result to $ax + by + cz + 6 = 0$, we identify $a = 11$, $b = -7$, and $c = 10$, which leads to $a + b + c = 14$.

The centroid of a triangle is defined as the arithmetic mean of its three vertices. Given the vertices $(2,4,6)$ and $(0,-2,-5)$ and the centroid $(2,1,-1)$, the coordinates of the third vertex $(x, y, z)$ satisfy the relations $2+0+x = 3(2)$, $4-2+y = 3(1)$, and $6-5+z = 3(-1)$. Solving these linear equations gives the third vertex as $C(4, 1, -4)$.

To find the image $(\alpha, \beta, \gamma)$ of this vertex in the plane $x+2y+4z=11$, we use the reflection formula which relates the point to its image across a plane. The coordinates satisfy the ratio:
$\frac{\alpha-4}{1} = \frac{\beta-1}{2} = \frac{\gamma+4}{4} = -2 \frac{4+2(1)+4(-4)-11}{1^{2+}2^{2+}4^2}$
Calculating the numerator, we get $4+2-16-11 = -21$, and the denominator is $1+4+16 = 21$. This simplifies the ratio to $-2 \times (-21)/21$, which equals $2$. Setting each component equal to this constant, we find $\alpha-4 = 2(1)$, $\beta-1 = 2(2)$, and $\gamma+4 = 2(4)$, which results in $\alpha=6$, $\beta=5$, and $\gamma=4$.

The final expression to evaluate is $\alpha\beta+\beta\gamma+\gamma\alpha$. Substituting our values gives $(6)(5)+(5)(4)+(4)(6)$, which simplifies to $30+20+24$, yielding 74.

To determine the equation of a plane perpendicular to a specific line, remember that the normal vector of the required plane is identical to the direction vector of that line. Since the line is defined by the intersection of two planes with normal vectors $(1, -3, 2)$ and $(4, -1, 1)$, its direction vector is their cross product, calculated as

. Using this normal vector $(-1, 7, 11)$ and the given point $(1, 1, 2)$, the equation of the plane follows the point-normal form $-1(x - 1) + 7(y - 1) + 11(z - 2) = 0$.

Expanding this expression gives $-x + 1 + 7y - 7 + 11z - 22 = 0$, which simplifies to $-x + 7y + 11z = 28$. Transforming this into the standard form $Ax + By + Cz = 1$ requires dividing both sides of the equation by 28, yielding $-\frac{1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1$. Here, the coefficients are $A = -\frac{1}{28}$, $B = \frac{7}{28}$, and $C = \frac{11}{28}$. Substituting these values into the expression $140(C - B + A)$ results in $140\left(\frac{11}{28} - \frac{7}{28} - \frac{1}{28}\right)$, which simplifies to $140 \times \frac{3}{28}$ or 15.

The family of planes passing through the intersection of the planes $4x - y + z = 10$ and $x + y - z = 4$ is represented by the linear combination $(4x - y + z - 10) + \lambda(x + y - z - 4) = 0$. Regrouping the terms, we define the new plane as $(4+\lambda)x + (-1+\lambda)y + (1-\lambda)z - (10+4\lambda) = 0$. Because the plane is rotated by $\pi/2$, the normal vector of the new plane must be perpendicular to the normal vector of the original plane, implying their dot product equals zero. Evaluating the dot product $4(4+\lambda) - 1(-1+\lambda) + 1(1-\lambda) = 0$ yields $16 + 4\lambda + 1 - \lambda + 1 - \lambda = 0$, which simplifies to $18 + 2\lambda = 0$, meaning $\lambda = -9$. Substituting this value back into the general equation results in the plane $5x + 10y - 10z - 26 = 0$. The perpendicular distance $\alpha$ from the point (2, 3, -4) to this plane is given by the formula $\alpha = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$, which calculates to $\alpha = \frac{|5(2) + 10(3) - 10(-4) - 26|}{\sqrt{5^2 + 10^2 + (-10)^2}}$. Simplifying the numerator gives 54 and the denominator is $\sqrt{225} = 15$, resulting in $\alpha = \frac{54}{15}$. Multiplying this value by 35 gives $35 \alpha = 35 \times \frac{54}{15}$, which simplifies to $7 \times 18$, yielding 126.

Finding the image $P$ of a point reflected across a plane relies on the geometric property that the line connecting the point and its image is perpendicular to the plane, and the midpoint of this segment lies on the plane. For a point $(x_1, y_1, z_1)$ and a plane $ax+by+cz+d=0$, the reflection $(x, y, z)$ is determined by the ratio $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^{2+}b^{2+}c^2}$. Substituting the coordinates $\left( \frac{5}{3}, \frac{5}{3}, \frac{8}{3} \right)$ and the plane equation $x-2y+z-2=0$ into this expression yields a displacement constant of $\frac{1}{3}$, which results in the coordinates of the image point $P$ being $(2, 1, 3)$.

Applying the distance formula between $P(2, 1, 3)$ and $Q(6, -2, \alpha)$ provides the equation $(6-2)^2 + (-2-1)^2 + (\alpha-3)^2 = 13^2$. This simplifies to $16 + 9 + (\alpha-3)^2 = 169$, which further reduces to $(\alpha-3)^2 = 144$. Since the condition $\alpha > 0$ must be satisfied, we take the positive root $\alpha - 3 = 12$, resulting in $\alpha = 15$.

A plane passing through the intersection of two given planes can be represented by the equation $P_1 + \lambda P_2 = 0$, which helps define the family of planes and allows us to isolate the specific member using geometric constraints. By combining $(2x+3y-z-2) + \lambda(x+2y+3z-6) = 0$, we find the normal vector is $(2+\lambda, 3+2\lambda, 3\lambda-1)$, and since the plane is perpendicular to $2x+y-z+1=0$, their dot product must vanish, yielding $\lambda = -8$. Substituting this value into our expression gives the plane $6x+13y+25z-46=0$, and calculating the perpendicular distance $d$ from the point $(-7, 1, 1)$ results in $d = \frac{|6(-7)+13(1)+25(1)-46|}{\sqrt{6^{2+}13^{2+}25^2}} = \frac{50}{\sqrt{830}}$. Squaring this value provides $d^2 = \frac{2500}{830}$, which reduces to $\frac{250}{83}$.

To determine where two lines intersect, we equate their parametric representations and search for common parameter values. We can express any point on the first line as $(x, y, z) = (\lambda+1, 2\lambda+2, \lambda-3)$ and any point on the second line as $(x, y, z) = (2\mu+a, 3\mu-2, \mu+3)$ by introducing parameters $\lambda$ and $\mu$. For these lines to intersect at a specific point $P$, these coordinates must be equal simultaneously. Equating the $z$-components gives $\lambda-3 = \mu+3$, which simplifies to $\lambda = \mu+6$. Equating the $y$-components gives $2\lambda+2 = 3\mu-2$, or $2\lambda = 3\mu-4$. Substituting the expression for $\lambda$ into the second equation yields $2(\mu+6) = 3\mu-4$, which simplifies further to $2\mu+12 = 3\mu-4$, confirming that $\mu = 16$ and $\lambda = 22$.

With the parameter $\lambda=22$ determined, the $z$-coordinate of the point $P$ is found by evaluating $\lambda-3$, which gives $19$. To find the constant $a$, we equate the $x$-coordinates using the value $\mu=16$, so $\lambda+1 = 2\mu+a$ becomes $22+1 = 2(16)+a$, leading to $23 = 32+a$ and $a = -9$. The problem now reduces to finding the distance of point $P$, which lies at height $z=19$, from the plane defined by $z=-9$. The vertical distance is the absolute difference between these values, calculated as $|19 - (-9)|$, which equals $28$.

To find the distance from a point to a plane, one must first establish the Cartesian equation of the plane defined by the three specified points $(2,-3,1),(-1,1,-2)$, and $(3,-4,2)$. By constructing vectors between these points and determining their normal vector, the equation of the plane simplifies to:
$x - z - 1 = 0$
Using the standard perpendicular distance formula, we substitute the coordinates $(7,-3,-4)$ into the expression to calculate the final result:
$d = \left| \frac{7+4-1}{\sqrt{2}} \right| = 5 \sqrt{2}$

The shortest distance between two skew lines in 3D space is obtained by projecting the vector connecting a point on each line onto the vector that is mutually perpendicular to both lines. The position vectors $\vec{a} = \hat{i} - 8\hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 6\hat{k}$ identify points on the two lines, while their respective directions are given by $\vec{p} = 2\hat{i} - 7\hat{j} + 5\hat{k}$ and $\vec{q} = 2\hat{i} + \hat{j} - 3\hat{k}$.

The vector connecting the lines is $\vec{b} - \vec{a} = 10\hat{j} + 2\hat{k}$, and the mutual perpendicular direction is found using the cross product:
$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = 16\hat{i} + 16\hat{j} + 16\hat{k}$
Applying these values to the distance formula:
$d = \frac{|(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|} = \frac{|(0)(16) + (10)(16) + (2)(16)|}{\sqrt{16^2 + 16^2 + 16^2}} = \frac{192}{16\sqrt{3}} = 4\sqrt{3}$

The family of planes passing through the intersection of two given planes is represented by the linear combination $P_1 + kP_2 = 0$, where $P_1$ and $P_2$ are the plane equations and $k$ is a scalar constant. Writing this for the given planes yields $(x + (\lambda + 4)y + z - 1) + k(2x + y + z - 2) = 0$. Since this plane passes through $(1, 0, 1)$, substituting these coordinates gives $1 + k(2 + 1 - 2) = 0$, which simplifies to $1 + k = 0$, leading to $k = -1$. Knowing $k = -1$, we substitute the other point $(0, 1, 0)$ into the combined equation to determine $\lambda$, resulting in $(0 + \lambda + 4 + 0 - 1) - 1(0 + 1 + 0 - 2) = 0$. This simplifies to $\lambda + 3 + 1 = 0$, which means $\lambda = -4$.

With $\lambda$ determined, the target point $(2\lambda, \lambda, -\lambda)$ becomes $(-8, -4, 4)$. The perpendicular distance of this point from the plane $2x + y + z - 2 = 0$ is determined by the formula $d = \frac{|ax_0 + by_0 + cz_0 + D|}{\sqrt{a^2 + b^2 + c^2}}$. Substituting these values yields $d = \frac{|2(-8) - 4 + 4 - 2|}{\sqrt{2^2 + 1^2 + 1^2}}$, which simplifies to $\frac{|-18|}{\sqrt{6}}$, ultimately resulting in $3\sqrt{6}$.

The equation of a plane containing the intersection of two lines is determined by the family of planes equation, where a specific member is found by ensuring the normal vector is orthogonal to the direction vector of a line parallel to the plane. The family of planes passing through the intersection of $2x+y-z-3=0$ and $5x-3y+4z+9=0$ is given by $(2x+y-z-3) + \lambda(5x-3y+4z+9) = 0$, which simplifies to $(5\lambda+2)x + (1-3\lambda)y + (4\lambda-1)z + (9\lambda-3) = 0$. Setting the dot product of the normal vector $(5\lambda+2, 1-3\lambda, 4\lambda-1)$ and the direction vector $(2, 4, 5)$ of the line $\frac{x+2}{2}=\frac{y-3}{4}=\frac{z-7}{5}$ to zero, we solve $2(5\lambda+2) + 4(1-3\lambda) + 5(4\lambda-1) = 0$, yielding $18\lambda + 3 = 0$ and $\lambda = -1/6$. Substituting this parameter results in the plane equation $7x+9y-10z-27=0$.

The distance from point $A(8,-1,-19)$ to this plane, measured along the line path with direction vector $(-3, 4, 12)$ derived from $\frac{x}{-3}=\frac{y-5}{4}=\frac{z-2}{12}$, is calculated by finding the intersection point $B$ between the plane and this path. Expressing the coordinates of point $B$ as $(8-3k, -1+4k, -19+12k)$ for a parameter $k$, we substitute these into the plane equation $7(8-3k) + 9(-1+4k) - 10(-19+12k) - 27 = 0$. Solving this linear equation leads to $56 - 21k - 9 + 36k + 190 - 120k - 27 = 0$, which simplifies to $-105k + 210 = 0$, giving $k = 2$. With the intersection point $B$ identified as $(2, 7, 5)$, the requested distance is the magnitude of the displacement vector from $A$ to $B$, computed as $\sqrt{(2-8)^2 + (7+1)^2 + (5+19)^2} = \sqrt{(-6)^2 + 8^2 + 24^2} = \sqrt{36+64+576} = \sqrt{676} = 26$.

The perpendicular distance from point $P(4, 6, -2)$ to a line requires identifying the foot of the perpendicular, denoted as $M$, on that line. Representing any point $M$ on the line with parameter $\lambda$, we obtain coordinates $(3\lambda-3, 3\lambda+2, 3-\lambda)$ because the line passes through $(-3, 2, 3)$ with direction ratios $(3, 3, -1)$. The vector $\vec{PM}$ is calculated by subtracting the coordinates of $P$ from those of $M$, resulting in $(3\lambda-7, 3\lambda-4, 5-\lambda)$. Since the segment $PM$ must be perpendicular to the line, the dot product of $\vec{PM}$ and the line's direction vector $(3, 3, -1)$ must be zero, leading to the following equation:

$3(3\lambda-7) + 3(3\lambda-4) - 1(5-\lambda) = 0$

Solving this expression results in $19\lambda - 38 = 0$, establishing that $\lambda = 2$. Substituting this value back into the expression for $M$ yields the coordinates $(3, 8, 1)$. Applying the distance formula between $P(4, 6, -2)$ and $M(3, 8, 1)$ gives the final calculation:

$\sqrt{(3-4)^2 + (8-6)^2 + (1-(-2))^2} = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$

The equation of a plane perpendicular to a line uses the direction vector of that line as its normal vector, which dictates the coefficients of the plane equation. We find the direction vector by subtracting the coordinates of the two given points, $(1, 2, 3)$ and $(-2, 3, 5)$, which yields $(-2 - 1, 3 - 2, 5 - 3)$ or $(-3, 1, 2)$. Using this normal vector, the equation of the plane takes the form $-3x + y + 2z = k$, which can be rewritten as $3x - y - 2z = -k$. By substituting the point $(3, -2, 5)$ into this expression, we obtain $3(3) - (-2) - 2(5) = 9 + 2 - 10 = 1$.

Comparing this result to the required form $\alpha x + \beta y + \gamma z = 1$, we identify the coefficients as $\alpha = 3$, $\beta = -1$, and $\gamma = -2$, noting that $\gamma$ replaces the third variable. Consequently, the value of the product $\alpha \beta \gamma$ is determined by calculating $3 \times (-1) \times (-2)$, which equals 6.

A plane is uniquely determined by a point it contains and a normal vector perpendicular to its surface. The vector connecting $(4, 1, -3)$ and $(2, 4, 3)$ serves as this normal, which is found by calculating the difference between the points to obtain $(-2, 3, 6)$. Using the vector $(2, -3, -6)$ simplifies the arithmetic, and applying the point-normal form $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ with the point $(1, -1, -5)$ results in the equation $2(x-1) - 3(y+1) - 6(z+5) = 0$.

Simplifying this expression yields $2x - 3y - 6z = 35$. The perpendicular distance of the point $(3, -2, 2)$ from this plane follows the formula $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}$. Substituting the coordinates $(3, -2, 2)$ and the coefficients from the plane equation into this expression provides $d = \frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{2^2 + (-3)^2 + (-6)^2}}$.

Evaluating the numerator gives $|6 + 6 - 12 - 35|$, which simplifies to $|-35|$, while the denominator becomes $\sqrt{4 + 9 + 36}$, which is $\sqrt{49}$ or 7. Dividing these values results in 5.