Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The distance of the point $(-1, 2, 3)$ from the plane $\vec{r} \cdot (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 10$ along the path of the shortest distance can be found by identifying the line that passes through the point in the direction of the common perpendicular to the two skew lines. This perpendicular direction is obtained by the cross product of the direction vectors $(2 \hat{i} + \hat{k})$ and $(\hat{i} - \hat{j} + \hat{k})$, which results in $\vec{n} = \hat{i} - \hat{j} - 2 \hat{k}$. Based on this direction, the line passing through $(-1, 2, 3)$ is defined by the equations:
$\frac{x+1}{1} = \frac{y-2}{-1} = \frac{z-3}{-2} = r$
Any point on this line can thus be represented by the coordinates $(r-1, 2-r, 3-2r)$.

The intersection point $Q$ with the plane $x - 2y + 3z = 10$ is found by substituting these coordinates into the plane equation, which gives $(r-1) - 2(2-r) + 3(3-2r) = 10$. Simplifying this leads to $r - 1 - 4 + 2r + 9 - 6r = 10$, or $-3r + 4 = 10$, yielding $r = -2$. Substituting $r = -2$ into the parametric expressions gives the coordinates $Q(-3, 4, 7)$. Using the distance formula between $P(-1, 2, 3)$ and $Q(-3, 4, 7)$ provides $\sqrt{(-3 - (-1))^2 + (4 - 2)^2 + (7 - 3)^2} = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{4+4+16} = 2\sqrt{6}$.

The angle between two planes is determined by the angle between their normal vectors, $\vec{n}_1 = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} - \hat{j} + \hat{k}$. The cosine of the angle $\theta$ between these planes is calculated by dividing the absolute value of the dot product of these vectors by the product of their magnitudes:
$\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|} = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^{2+}1^{2+}2^2} \sqrt{2^{2+}(-1)^{2+}1^2}} = \frac{3}{6} = \frac{1}{2}$
This calculation establishes that $\theta = \frac{\pi}{3}$.

Because the line $L$ forms an angle $\theta$ with the normal vector of the plane $P_2$, the angle $\alpha$ between the line and the plane $P_2$ is the complement of $\theta$, defined by the relation $\alpha = \frac{\pi}{2} - \theta$. Substituting the value of $\theta$, we find $\alpha = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$. The required value of the expression $(\tan^2 \theta) (\cot^2 \alpha)$ is determined by substituting these angles:
$(\tan^2 \theta) (\cot^2 \alpha) = \left(\tan \frac{\pi}{3}\right)^2 \left(\cot \frac{\pi}{6}\right)^2 = 3 \times 3 = 9$

The angle between a line and a plane is determined by the relationship between the line's direction vector $\mathbf{v}$ and the plane's normal vector $\mathbf{n}$, using the sine of the angle $\theta$ given by $\sin \theta = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}| |\mathbf{n}|}$. The line $\ell$, given as $x = \frac{1-y}{-2} = \frac{z-3}{\lambda}$, is rewritten in standard form as $\frac{x}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda}$, identifying the direction vector as $\mathbf{v} = (1, 2, \lambda)$. The plane $x+2y+3z=4$ has a normal vector $\mathbf{n} = (1, 2, 3)$. Given $\cos \theta = \sqrt{\frac{5}{14}}$, the value of $\sin \theta$ is $\sqrt{1 - \frac{5}{14}} = \frac{3}{\sqrt{14}}$. Setting the geometric formula equal to this value, we have $\frac{3}{\sqrt{14}} = \frac{|(1)(1) + (2)(2) + (\lambda)(3)|}{\sqrt{1^{2+}2^{2+}\lambda^2} \sqrt{1^{2+}2^{2+}3^2}} = \frac{|5+3\lambda|}{\sqrt{5+\lambda^2} \sqrt{14}}$. Squaring both sides of $3\sqrt{5+\lambda^2} = |5+3\lambda|$ results in $9(5+\lambda^2) = (5+3\lambda)^2$, which expands to $45+9\lambda^2 = 25+30\lambda+9\lambda^2$, yielding $\lambda = \frac{2}{3}$.

Any arbitrary point on the line can be represented in terms of a parameter $t$ as $(t, 2t+1, \frac{2}{3}t+3)$. Substituting these coordinates into the plane equation $x+2y+3z=4$ gives $t + 2(2t+1) + 3(\frac{2}{3}t+3) = 4$. Expanding this expression results in $t + 4t + 2 + 2t + 9 = 4$, which simplifies to $7t + 11 = 4$, leading to $7t = -7$ and thus $t = -1$. The intersection point $(\alpha, \beta, \gamma)$ is therefore $(-1, 2(-1)+1, \frac{2}{3}(-1)+3)$, which is $(-1, -1, \frac{7}{3})$. Evaluating the final expression $\alpha+2\beta+6\gamma$ with these values gives $-1 + 2(-1) + 6(\frac{7}{3}) = -3 + 14 = 11$.

To determine the equation of a plane that is parallel to two intersecting lines, we first identify the normal vector to the plane, which must be perpendicular to the direction vectors of both lines, $\langle 1, 2, 3 \rangle$ and $\langle 1, 1, 5 \rangle$. By calculating the cross product of these direction vectors, we obtain

, providing the normal components $7, -2, -1$. We find the intersection point $S$ by equating the parametric coordinates of the two lines, which yields $\lambda = 2$ and $\mu = 1$, corresponding to the point $(2, 4, 6)$.

Using the point-normal form of a plane equation, $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$, we substitute our normal components and the intersection point to get $7(x - 2) - 2(y - 4) - 1(z - 6) = 0$. Expanding this expression results in $7x - 14 - 2y + 8 - z + 6 = 0$, which simplifies to $7x - 2y - z = 0$. Comparing this equation to the given form $ax + by - z + d = 0$, we identify $a = 7$, $b = -2$, and $d = 0$. Therefore, the value of the sum is $a + b + d = 5$.

The shortest distance between two skew lines defined by $\vec{r} = \vec{a_1} + t\vec{p}$ and $\vec{r} = \vec{a_2} + s\vec{q}$ is calculated using the formula $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$. Based on the given equations, we identify the position vectors as $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$, alongside the direction vectors $\vec{p} = 2\hat{i} + 3\hat{j} + \lambda\hat{k}$ and $\vec{q} = \hat{i} + 4\hat{j} + 5\hat{k}$. Calculating the difference between the position vectors yields $\vec{a_2} - \vec{a_1} = \hat{i} + 2\hat{j} + 2\hat{k}$, while the cross product of the direction vectors is $(15 - 4\lambda)\hat{i} - (10 - \lambda)\hat{j} + 5\hat{k}$.

Substituting these values into the distance formula and setting the result to $\frac{1}{\sqrt{3}}$, we determine that the numerator, derived from the dot product $|(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})|$, simplifies to $|(1)(15 - 4\lambda) - 2(10 - \lambda) + 2(5)|$, which evaluates to $|5 - 2\lambda|$. Squaring both sides of the resulting equation $\left| \frac{5 - 2\lambda}{\sqrt{(15 - 4\lambda)^2 + (10 - \lambda)^2 + 25}} \right| = \frac{1}{\sqrt{3}}$ produces $3(5 - 2\lambda)^2 = (15 - 4\lambda)^2 + (10 - \lambda)^2 + 25$. Expanding and simplifying this expression results in the quadratic equation $5\lambda^2 - 80\lambda + 275 = 0$. By applying the relation for the sum of roots, which is the negative of the ratio of the linear coefficient to the leading coefficient, we find that the sum of all possible values for $\lambda$ is $\frac{80}{5}$, equaling 16.

The family of planes passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ is given by the equation $(x+2y+az-2) + \lambda(x-y+z-3) = 0$. Rearranging this expression by grouping the coefficients of $x$, $y$, and $z$, we obtain $(1+\lambda)x + (2-\lambda)y + (a+\lambda)z = 2+3\lambda$. Since this equation represents the same plane as $5x - 11y + bz = 6a-1$, the coefficients must be proportional, resulting in the ratios $\frac{1+\lambda}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1}$.

Solving the first equality $\frac{1+\lambda}{5} = \frac{2-\lambda}{-11}$ leads to $-11-11\lambda = 10-5\lambda$, which simplifies to $6\lambda = -21$, or $\lambda = -3.5$. Substituting this value into the proportionality ratio, we find the constant of proportionality to be $\frac{1-3.5}{5} = -0.5$. Setting the final ratio equal to this constant gives $\frac{2+3(-3.5)}{6a-1} = -0.5$, which reduces to $\frac{-8.5}{6a-1} = -0.5$, or $6a-1 = 17$, yielding $a=3$. Substituting $a=3$ into $\frac{a+\lambda}{b} = -0.5$ gives $\frac{3-3.5}{b} = -0.5$, which confirms $b=1$.

With $a=3$ and $b=1$, the plane equation is $5x - 11y + z = 17$. The distance from the point $(3, -c, c)$ to this plane is given by the formula $\frac{|5(3) - 11(-c) + 1(c) - 17|}{\sqrt{5^2 + (-11)^2 + 1^2}}$, which simplifies to $\frac{|12c-2|}{\sqrt{147}}$, or $\frac{|12c-2|}{7\sqrt{3}}$. Equating this to the given distance $\frac{2}{\sqrt{3}}$ allows us to write $\frac{|12c-2|}{7\sqrt{3}} = \frac{2}{\sqrt{3}}$, which implies $|12c-2| = 14$. For $c \in \mathbb{Z}$, the equation $12c-2 = -14$ yields $c=-1$. Calculating the requested value $\frac{a+b}{c}$ gives $\frac{3+1}{-1}$, which is -4.

The shortest distance between two skew lines can be determined using the formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$, where $\vec{a}_1$ and $\vec{a}_2$ are position vectors of points on each line, and $\vec{b}_1$ and $\vec{b}_2$ are their respective direction vectors. For line $L$, we have the point $(5, \lambda, -\lambda)$ and the direction vector $\vec{b}_1 = (-2, 0, 1)$. For line $L_1$, expressed in symmetric form as $\frac{x+1}{1} = \frac{y-1}{1} = \frac{z-4}{-1}$, the point is $(-1, 1, 4)$ and the direction vector is $\vec{b}_2 = (1, 1, -1)$.

Calculating the cross product of the direction vectors yields

, which has a magnitude of $\sqrt{(-1)^2 + (-1)^2 + (-2)^2} = \sqrt{6}$. Using the position vectors $\vec{a}_1 = (5, \lambda, -\lambda)$ and $\vec{a}_2 = (-1, 1, 4)$, the difference $\vec{a}_2 - \vec{a}_1$ is $(-6, 1-\lambda, 4+\lambda)$. Substituting these into the distance formula gives $2\sqrt{6} = \frac{|(-6)(-1) + (1-\lambda)(-1) + (4+\lambda)(-2)|}{\sqrt{6}}$, which simplifies to $|6 - 1 + \lambda - 8 - 2\lambda| = 12$, or $|-3 - \lambda| = 12$. This results in $\lambda = 9$ or $\lambda = -15$. Given the constraint $\lambda \geq 0$, we must have $\lambda = 9$.

With $\lambda = 9$, the line $L$ becomes $\frac{x-5}{-2} = \frac{y-9}{0} = \frac{z+9}{1} = k$. Any point $(\alpha, \beta, \gamma)$ on this line can be written in terms of parameter $k$ as $\alpha = 5 - 2k$, $\beta = 9$, and $\gamma = k - 9$. Evaluating the expression $\alpha + 2\gamma$ for any point on the line results in $(5 - 2k) + 2(k - 9) = 5 - 2k + 2k - 18 = -13$. Since the expression $\alpha + 2\gamma$ must always equal $-13$, the equation $\alpha + 2\gamma = 24$ is not possible.

To find the area of triangle $PQR$, we recognize that $P$ is a fixed point and $Q$ and $R$ are points on a line equidistant from $P$. This configuration forms an isosceles triangle where the altitude dropped from $P$ to the line intersects the midpoint of the base $QR$. Let $T$ be the foot of the perpendicular from $P(4, 2, 7)$ onto the line given by $\frac{x+1}{2} = \frac{y+2}{3} = \frac{z-1}{2}$. Any point on this line can be represented as $(2t-1, 3t-2, 2t+1)$ for some scalar $t$. The direction vector of the line is $\vec{d} = (2, 3, 2)$, and the vector $\vec{PT} = (2t-1-4, 3t-2-2, 2t+1-7) = (2t-5, 3t-4, 2t-6)$ must be perpendicular to $\vec{d}$.

Setting the dot product of $\vec{PT}$ and $\vec{d}$ to zero, we obtain $2(2t-5) + 3(3t-4) + 2(2t-6) = 0$. Expanding this leads to $4t-10 + 9t-12 + 4t-12 = 0$, or $17t - 34 = 0$, which yields $t=2$. Substituting $t=2$ back into the coordinates gives $T(3, 4, 5)$. The length of the altitude $PT$ is calculated as the distance between $P(4, 2, 7)$ and $T(3, 4, 5)$, resulting in $\sqrt{(3-4)^2 + (4-2)^2 + (5-7)^2} = \sqrt{1+4+4} = 3$. In the right-angled triangle $PQT$, where $PQ = \sqrt{26}$ is the hypotenuse and $PT = 3$ is the altitude, the length $QT$ is $\sqrt{26 - 3^2} = \sqrt{17}$.

Because $T$ serves as the midpoint of the base $QR$, the length of the base is $QR = 2QT = 2\sqrt{17}$. The area of triangle $PQR$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{17} \times 3 = 3\sqrt{17}$. Squaring this area provides the final result of $(3\sqrt{17})^2 = 9 \times 17 = 153$.

The shortest distance between two skew lines in space corresponds to the magnitude of the projection of the vector joining any two points on the lines onto the common perpendicular of the direction vectors. The line $I_1$ lies in the $xy$-plane with intercepts $1/8$ and $1/(4\sqrt{2})$, which means it passes through $(1/8, 0, 0)$ and $(0, 1/(4\sqrt{2}), 0)$, yielding a direction vector $\vec{b}_1 = (1, -\sqrt{2}, 0)$. The line $I_2$ is in the $zx$-plane with intercepts $-1/8$ and $-1/(6\sqrt{3})$, passing through $(-1/8, 0, 0)$ and $(0, 0, -1/(6\sqrt{3}))$, which results in a direction vector $\vec{b}_2 = (-6\sqrt{3}, 0, 8)$.

Computing the common perpendicular vector via the cross product gives $\vec{n} = \vec{b}_1 \times \vec{b}_2 = (-8\sqrt{2}, -8, -6\sqrt{6})$. With points $A_1 = (1/8, 0, 0)$ and $A_2 = (-1/8, 0, 0)$, the displacement vector between them is $\vec{v} = (-1/4, 0, 0)$. The shortest distance is then determined using the expression:
$d = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{n}|} = \frac{|(-1/4, 0, 0) \cdot (-8\sqrt{2}, -8, -6\sqrt{6})|}{\sqrt{(-8\sqrt{2})^2 + (-8)^2 + (-6\sqrt{6})^2}}$

Simplifying this calculation produces $d = \frac{2\sqrt{2}}{\sqrt{128 + 64 + 216}} = \frac{2\sqrt{2}}{\sqrt{408}} = \frac{2\sqrt{2}}{2\sqrt{102}} = \frac{1}{\sqrt{51}}$. Squaring the reciprocal of this distance results in $d^{-2} = (\sqrt{51})^2 = 51$.

The direction of a line perpendicular to two given lines is determined by the cross product of their direction vectors, and since this line must pass through the origin, the cross product directly defines its orientation. We identify the direction vectors $\vec{v_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{v_2} = 2\hat{i} + 2\hat{j} + \hat{k}$. Their cross product $\vec{d} = \vec{v_1} \times \vec{v_2}$ is calculated as $\hat{i}(2-6) - \hat{j}(1-6) + \hat{k}(2-4)$, which results in $\vec{d} = -4\hat{i} + 5\hat{j} - 2\hat{k}$. Given that the line passes through the origin, its parametric equation is $x = -4t, y = 5t, z = -2t$.

The intersection point $P$ lies on this line and $l_1: \vec{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$. Equating their coordinates leads to the system $-4t = 1+\lambda$, $5t = -11+2\lambda$, and $-2t = -7+3\lambda$. Solving this system, where the first two equations imply $\lambda = -4t-1$ and $5t = -11+2(-4t-1)$, yields $t = -1$, which confirms the intersection point as $P(4, -5, 2)$.

For the point $Q$ on $l_2$, any arbitrary point is represented as $Q(-1+2\mu, 2\mu, 1+\mu)$. The vector $\vec{PQ}$ connecting $P(4, -5, 2)$ to $Q$ is $\vec{PQ} = (-1+2\mu-4)\hat{i} + (2\mu+5)\hat{j} + (1+\mu-2)\hat{k}$, which simplifies to $(2\mu-5)\hat{i} + (2\mu+5)\hat{j} + (\mu-1)\hat{k}$. Since $Q$ is the foot of the perpendicular from $P$ onto $l_2$, $\vec{PQ}$ must be orthogonal to the direction vector of $l_2$, $(2, 2, 1)$, requiring their dot product to be zero: $2(2\mu-5) + 2(2\mu+5) + 1(\mu-1) = 0$. Solving this results in $4\mu - 10 + 4\mu + 10 + \mu - 1 = 0$, leading to $9\mu = 1$, or $\mu = \frac{1}{9}$. The coordinates of $Q$ are $(-\frac{7}{9}, \frac{2}{9}, \frac{10}{9})$, and the final expression $9(\alpha+\beta+\gamma)$ evaluates to $9(-\frac{7}{9} + \frac{2}{9} + \frac{10}{9}) = 5$.

The area of a triangle can be expressed as half the product of its base and the perpendicular height from the opposite vertex. Given that the area of $\triangle ABC$ is 21 sq. units and the base segment $BC$ measures $2\sqrt{21}$ units, we can determine the perpendicular distance $h$ from the vertex $A(0,2,\alpha)$ to the line $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$ by substituting these values into the area formula $\frac{1}{2} \times 2\sqrt{21} \times h = 21$, which simplifies to $h = \sqrt{21}$.

To calculate this height, consider a point $P(-\alpha, 1, -4)$ on the line, where the direction vector is $\vec{v} = (5, 2, 3)$. The perpendicular distance from point $A$ to the line is given by the magnitude of the cross product of the vector $\vec{AP}$ and the direction vector $\vec{v}$, divided by the magnitude of $\vec{v}$. Here, $\vec{AP} = (-\alpha, -1, -\alpha-4)$, and the cross product $\vec{AP} \times \vec{v}$ yields the vector $(2\alpha+5, -(2\alpha+20), 5-2\alpha)$. Squaring both sides of the distance relation $h^2 = \frac{|\vec{AP} \times \vec{v}|^2}{|\vec{v}|^2}$ results in the equation $\frac{(2\alpha+5)^{2+}(2\alpha+20)^{2+}(5-2\alpha)^2}{38} = 21$.

Expanding this leads to $12\alpha^{2+}80\alpha+450 = 798$, which simplifies to the quadratic equation $12\alpha^{2+}80\alpha-348 = 0$. Solving this for the integer $\alpha$ yields $\alpha = 3$, and consequently, $\alpha^2 = 9$.

The equation of a plane is determined by its normal vector and a point through which it passes. Because the required plane contains the line formed by the intersection of $x+2 y+3 z-4=0$ and $2 x+y-z+5=0$, the normal vector of the plane must be perpendicular to the direction vector of this line. This direction vector, $\vec{n}_1 = -5\hat{i}+7\hat{j}-3\hat{k}$, is calculated by taking the cross product of the normal vectors of the two given planes. Similarly, since the target plane is perpendicular to the second plane, its normal must also be perpendicular to the normal vector of that second plane, which is $\vec{n}_2 = 5\hat{i}-2\hat{j}-3\hat{k}$.

By taking the cross product of $\vec{n}_1$ and $\vec{n}_2$, we find the normal to the required plane to be $-27\hat{i}-30\hat{j}-25\hat{k}$, leading to the general equation:
$27x + 30y + 25z = 4$
Substituting the point $(0, -\frac{11}{5}, \frac{14}{5})$, which lies on the line of intersection, confirms the value of the constant term. With the coefficients $a=27$, $b=30$, and $c=25$, the final calculation of $a-b+c$ yields $27 - 30 + 25$, which equals 22.

Two lines intersect at a common point if there exist parameters $\lambda$ and $\mu$ such that their corresponding coordinates are identical. Comparing the $y$-components, the equation $1-\lambda=0$ reveals that $\lambda=1$. Similarly, the $z$-components must satisfy $-1=-1+\mu a$, which simplifies to $\mu a=0$; since the problem states $a \neq 0$, it implies that $\mu=0$. Substituting these values into the $x$-components yields $1+\lambda a=4+2\mu$, which becomes $1+a=4$, determining that $a=3$. With these parameters, the intersection point is $(4, 0, -1)$. The square of the distance of this point from the origin is calculated as:
$d^2 = 4^2 + 0^2 + (-1)^2 = 17$

The angle $\theta$ between a line with direction vector $\vec{v}$ and a plane with normal vector $\vec{n}$ is determined by the relation $\sin\theta = \frac{|\vec{n} \cdot \vec{v}|}{|\vec{n}| |\vec{v}|}$. Given the plane $ax+y-z=b$, the normal vector is $\vec{n} = (a, 1, -1)$. The line defined by $x-1=a-y=z+1$ can be rewritten as $\frac{x-1}{1} = \frac{y-a}{-1} = \frac{z+1}{1}$, which provides the direction vector $\vec{v} = (1, -1, 1)$.

Given $\cos\theta = \frac{1}{3}$, the value of $\sin\theta$ is $\sqrt{1 - \left(\frac{1}{3}\right)^2} = \frac{2\sqrt{2}}{3}$. Substituting the vectors into the angular formula yields $\frac{|a(1) + 1(-1) + (-1)(1)|}{\sqrt{a^{2+}1^{2+}(-1)^2}\sqrt{1^{2+}(-1)^{2+}1^2}} = \frac{2\sqrt{2}}{3}$, which simplifies to $\frac{|a-2|}{\sqrt{3(a^{2+}2)}} = \frac{2\sqrt{2}}{3}$. Squaring both sides produces $\frac{(a-2)^2}{3(a^{2+}2)} = \frac{8}{9}$, which further reduces to $3(a^{2-}4a+4) = 8(a^{2+}2)$ or $5a^{2+}12a+4=0$. This quadratic equation factors as $(5a+2)(a+2)=0$, giving $a=-\frac{2}{5}$ or $a=-2$; since $a \in \mathbb{Z}$, we must have $a=-2$.