Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

A line parallel to a given plane must have a direction vector orthogonal to the plane's normal, $\vec{n} = \hat{i} + 2\hat{j} - \hat{k}$. Furthermore, because the line intersects the given line $L_1$, which passes through $(-1, 3, 2)$ with a direction vector $-3\hat{i} + 2\hat{j} - \hat{k}$, the required line must lie within the plane containing both $L_1$ and the point $(-4, 3, 1)$. We find the normal to this second plane by calculating the cross product of the direction of $L_1$ and the vector connecting the point $(-4, 3, 1)$ to $(-1, 3, 2)$, resulting in

. The direction vector of the required line is then obtained by taking the cross product of these two normal vectors,

, which simplifies to $3\hat{i} - \hat{j} + \hat{k}$. Utilizing this direction vector alongside the point $(-4, 3, 1)$, we obtain the equation $\frac{x+4}{3} = \frac{y-3}{-1} = \frac{z-1}{1}$.

Defining a plane in three-dimensional space requires a normal vector and a point that satisfies the planar equation. Since the required plane is perpendicular to the plane $2x - 5y = 15$, its normal vector must be orthogonal to the direction $(2, -5, 0)$, and because it contains the points $(3, 4, 2)$ and $(7, 0, 6)$, the normal vector must also be orthogonal to the vector connecting these two points, which is $(7-3, 0-4, 6-2) = (4, -4, 4)$. Computing the cross product of $(2, -5, 0)$ and $(4, -4, 4)$ yields the vector $-20\hat{i} - 8\hat{j} + 12\hat{k}$, which simplifies to the direction ratios $(5, 2, -3)$.

With the normal vector identified as $(5, 2, -3)$, the equation of the plane takes the form $5x + 2y - 3z = D$. Substituting the coordinates $(7, 0, 6)$ into this expression gives $5(7) + 2(0) - 3(6) = 35 - 18 = 17$, establishing the complete equation $5x + 2y - 3z = 17$. Because the point $(2, \alpha, \beta)$ lies on this plane, substituting these variables results in $5(2) + 2\alpha - 3\beta = 17$. Subtracting $10$ from both sides directly yields $2\alpha - 3\beta = 7$.

When three points lie on the same line, the direction ratios between any two pairs of points must be proportional. Comparing the vector between $P(2, -3, 4)$ and $Q(8, 0, 10)$ with the vector between $P(2, -3, 4)$ and $R(4, y, z)$, we establish the following proportionality:

$\frac{4-2}{8-2} = \frac{y+3}{3} = \frac{z-4}{6}$

This expression simplifies to:

$\frac{2}{6} = \frac{y+3}{3} = \frac{z-4}{6}$

From $\frac{2}{6} = \frac{y+3}{3}$, we find $y+3=1$, which leads to $y=-2$. Similarly, from $\frac{2}{6} = \frac{z-4}{6}$, we find $z-4=2$, which leads to $z=6$. With the coordinates of $R$ determined as $(4, -2, 6)$, the distance from the origin is calculated using the square root of the sum of squares, resulting in $\sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14}$.

The area of a triangle is determined by the formula involving half the product of the base length and the altitude drawn to that base. Since the base $BC$ lies on a specific line and has a fixed length of 5 units, finding the shortest perpendicular distance from vertex $A(1, -1, 2)$ to this line provides the altitude required to compute the area.

To identify this altitude, consider any point $P$ on the line represented by the parametric coordinates $(3\lambda-2, 1, 4\lambda)$. We define the vector $\vec{AP}$ as $(3\lambda-3, 2, 4\lambda-2)$ and require it to be perpendicular to the direction vector of the line, which is $(3, 0, 4)$. Setting the dot product of these vectors to zero, we solve the equation $3(3\lambda-3) + 0(2) + 4(4\lambda-2) = 0$. This simplifies to $9\lambda - 9 + 16\lambda - 8 = 0$, leading to $25\lambda = 17$, or $\lambda = \frac{17}{25}$. Substituting this value back into the expression for $\vec{AP}$ gives the vector $(-\frac{24}{25}, 2, \frac{18}{25})$. The magnitude of this vector, which represents the altitude $h$, is calculated as $\sqrt{(-\frac{24}{25})^2 + 2^2 + (\frac{18}{25})^2} = \frac{2\sqrt{34}}{5}$. Using the area formula $\frac{1}{2} \times \text{base} \times \text{height}$, the calculation becomes $\frac{1}{2} \times 5 \times \frac{2\sqrt{34}}{5}$, which simplifies to $\sqrt{34}$.

The equation of any plane passing through the line of intersection of the planes $x+y+z-6=0$ and $2x+3y+z+5=0$ can be expressed as a linear combination of the two given equations, written in the form $(x+y+z-6) + \lambda(2x+3y+z+5) = 0$. By grouping the terms associated with each variable, this equation becomes $(1+2\lambda)x + (1+3\lambda)y + (1+\lambda)z + (5\lambda-6) = 0$. Since this plane is perpendicular to the $xy$-plane, its normal vector must be parallel to the $xy$-plane, meaning the coefficient of $z$ must be zero; setting $1+\lambda = 0$ gives $\lambda = -1$. Substituting $\lambda = -1$ back into the family equation simplifies it to $-x - 2y - 11 = 0$, which is equivalent to $x+2y+11=0$. The perpendicular distance from the point $(0,0,256)$ to this plane is found using the distance formula $d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^{2+}B^{2+}C^2}}$, which calculates as $\frac{|1(0)+2(0)+11|}{\sqrt{1^{2+}2^2}}$, resulting in the distance $\frac{11}{\sqrt{5}}$.

Any plane passing through the line of intersection of two given planes can be expressed as a linear combination of their equations, written in the form $(2x - y - 4) + \lambda(y + 2z - 4) = 0$. Because the required plane must pass through the specific point $(1, 1, 0)$, substituting these coordinates into the family of planes equation allows for the determination of the parameter $\lambda$. This substitution results in $(2(1) - 1 - 4) + \lambda(1 + 2(0) - 4) = 0$, which simplifies to $-3 + \lambda(-3) = 0$, confirming that $\lambda = -1$. Substituting $\lambda = -1$ back into the original linear combination yields $(2x - y - 4) - 1(y + 2z - 4) = 0$, which reduces to $2x - 2y - 2z = 0$, and dividing by $2$ simplifies the final equation to $x - y - z = 0$.

A plane is geometrically defined by its normal vector and its orientation relative to other surfaces. The equation of a plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ can be expressed in intercept form as $\frac{x}{a} - y + z = 1$, where the normal vector is $n_1 = (\frac{1}{a}, -1, 1)$. Given the reference plane $y - z + 5 = 0$ has a normal vector $n_2 = (0, 1, -1)$, we determine the plane's unknown parameter by relating their normals through the angle $\theta = \frac{\pi}{4}$ in the following formula:

$\cos \theta = \frac{|n_1 \cdot n_2|}{|n_1| |n_2|}$

Substituting the vector components gives $\frac{1}{\sqrt{2}} = \frac{|0 - 1 - 1|}{\sqrt{\frac{1}{a^2} + 2} \cdot \sqrt{2}}$. This expression simplifies to $\sqrt{\frac{1}{a^2} + 2} = 2$. Squaring both sides yields $\frac{1}{a^2} + 2 = 4$, which determines that $a = \pm \frac{1}{\sqrt{2}}$. These values yield the potential plane equations $\sqrt{2}x - y + z = 1$ and $\sqrt{2}x + y - z = -1$. Evaluating the point $(\sqrt{2}, 1, 4)$ in the second equation gives $\sqrt{2}(\sqrt{2}) + 1 - 4 = 2 + 1 - 4 = -1$. Since the result matches the equation, this point lies on the required plane.

The perpendicular distance from a point $P(\beta, 0, \beta)$ to the line $\frac{x}{1} = \frac{y-1}{0} = \frac{z+1}{-1}$ is determined by finding the point $Q(k, 1, -1-k)$ on the line such that the vector $\vec{PQ}$ is perpendicular to the line's direction vector $\vec{d} = (1, 0, -1)$. The condition $\vec{PQ} \cdot \vec{d} = 0$ simplifies as follows:
$(k-\beta)(1) + (1-0)(0) + (-1-k-\beta)(-1) = 0 \implies 2k + 1 = 0 \implies k = -1/2$
Substituting $k = -1/2$ identifies the point on the line as $Q(-1/2, 1, -1/2)$. The distance calculation from $P$ to $Q$ is given by:
$\sqrt{(\beta + 1/2)^2 + (0 - 1)^2 + (\beta + 1/2)^2} = \sqrt{3/2}$
$2(\beta + 1/2)^2 + 1 = 3/2 \implies 2(\beta + 1/2)^2 = 1/2$
$(\beta + 1/2)^2 = 1/4 \implies \beta + 1/2 = \pm 1/2$
This results in $\beta = 0$ or $\beta = -1,$ and given the constraint $\beta \neq 0,$ the required value is $\beta = -1$.

The equation of any plane passing through the intersection of the two planes $x+y+z=1$ and $2x+3y+4z=5$ is given by the linear combination $(x+y+z-1) + \lambda(2x+3y+4z-5) = 0$ Collecting the coefficients of $x$, $y$, and $z$, we obtain $(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda) = 0$. Since this new plane must be perpendicular to the plane $x - y + z = 0$, the dot product of their normal vectors must be zero. Computing this dot product, we have $(1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0$, which simplifies to $1+3\lambda = 0$, or $\lambda = -1/3$. Substituting $\lambda = -1/3$ back into the combined plane equation and multiplying by $3$ to clear the fraction yields $3(x+y+z-1) - (2x+3y+4z-5) = 0$, which reduces to $x - z + 2 = 0$. Converting this result to vector form using the position vector $\vec{r}$, we arrive at $\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$.

The angle $\theta$ between two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is determined by the formula
$\cos \theta = \frac{|l_1 l_2 + m_1 m_2 + n_1 n_2|}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$
Substituting $l = -3m - 5n$ from the given linear equation into the homogeneous quadratic equation $5lm - 2mn + 6nl = 0$ yields $5(-3m-5n)m - 2mn + 6n(-3m-5n) = 0$, which simplifies to $m^2 + 3mn + 2n^2 = 0$. Factoring this quadratic expression gives $(m+n)(m+2n)=0$, leading to two distinct relations for the direction ratios: $m = -n$ and $m = -2n$. When $m = -n$, substituting into $l = -3m - 5n$ yields $l = -2n$, resulting in direction ratios $(-2, -1, 1)$, and when $m = -2n$, substituting yields $l = n$, resulting in direction ratios $(1, -2, 1)$. Applying these sets of direction ratios to the angle formula gives
$\cos \theta = \frac{|(-2)(1) + (-1)(-2) + (1)(1)|}{\sqrt{(-2)^2 + (-1)^2 + 1^2} \sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|-2 + 2 + 1|}{\sqrt{6}\sqrt{6}} = \frac{1}{6}$
The angle between the lines is $\theta = \cos^{-1}\left(\frac{1}{6}\right)$.

The angle $\theta$ between a plane with normal vector $\vec{n}$ and a line with direction vector $\vec{b}$ is determined by the formula $\sin \theta = \frac{|\vec{n} \cdot \vec{b}|}{|\vec{n}| |\vec{b}|}$. The direction vector $\vec{b}$ of the line formed by the intersection of the planes $3x+4y+z-1=0$ and $5x+8y+2z+14=0$ is the cross product of their normal vectors $\vec{n}_1 = 3\hat{i} + 4\hat{j} + \hat{k}$ and $\vec{n}_2 = 5\hat{i} + 8\hat{j} + 2\hat{k}$, calculated as

. Given the normal vector of the plane $x+y+z=5$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$, substituting these vectors into the sine formula yields $\sin \theta = \frac{|(1)(0) + (1)(-1) + (1)(4)|}{\sqrt{1^{2+}1^{2+}1^2} \sqrt{0^{2+}(-1)^{2+}4^2}} = \frac{|3|}{\sqrt{3} \sqrt{17}} = \frac{3}{\sqrt{51}} = \frac{\sqrt{3}}{\sqrt{17}} = \sqrt{\frac{3}{17}}$. Therefore, the angle is $\sin^{-1}\left(\sqrt{\frac{3}{17}}\right)$.

The plane containing two intersecting lines is determined by a point lying on either line and a normal vector perpendicular to both lines, which is found by taking the cross product of their respective direction vectors. To find the direction vectors of $L_1$ and $L_2$, we compute the cross product of the normal vectors of the planes defining each line. The planes for $L_1$ have normal vectors $(2, -2, 3)$ and $(1, -1, 1)$, so its direction vector is $(2, -2, 3) \times (1, -1, 1) = (1, 1, 0)$. For $L_2$, the normal vectors are $(1, 2, -1)$ and $(3, -1, 2)$, yielding a direction vector of $(1, 2, -1) \times (3, -1, 2) = (3, -5, -7)$.

The normal vector to the required plane is the cross product of these two direction vectors, $(1, 1, 0) \times (3, -5, -7) = (-7, 7, -8)$. To find a point on the plane, we determine a point on $L_1$ by setting $y=0$ in its defining equations, $2x+3z-2=0$ and $x+z+1=0$, which solves to $x=-5$ and $z=4$, providing the point $(-5, 0, 4)$. Constructing the plane equation with normal $(-7, 7, -8)$ and point $(-5, 0, 4)$ yields $-7(x+5) + 7(y-0) - 8(z-4) = 0$, which simplifies to $-7x + 7y - 8z - 3 = 0$. The distance of the origin from this plane is calculated as follows:

$d = \frac{|-3|}{\sqrt{(-7)^2 + 7^2 + (-8)^2}} = \frac{3}{\sqrt{162}} = \frac{1}{3\sqrt{2}}$

The cosine of the angle $\theta$ between two lines with direction vectors $\mathbf{v}_1 = (a_1, b_1, c_1)$ and $\mathbf{v}_2 = (a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$. The first line has direction ratios $(2, 2, 1)$, while rearranging the second line to standard form $\frac{x-5}{2} = \frac{y-2}{p/7} = \frac{z-3}{4}$ provides direction ratios $(2, p/7, 4)$. Substituting these into the angle formula with $\cos \theta = \frac{2}{3}$ gives $\frac{|(2)(2) + (2)(p/7) + (1)(4)|}{\sqrt{2^{2+}2^{2+}1^2} \sqrt{2^{2+}(p/7)^{2+}4^2}} = \frac{2}{3}$. Simplifying the denominator leads to $\frac{|8 + 2p/7|}{3 \sqrt{20 + p^2/49}} = \frac{2}{3}$, which reduces to $\frac{|8 + 2p/7|}{\sqrt{20 + p^2/49}} = 2$. Squaring both sides results in $(8 + \frac{2p}{7})^2 = 4(20 + \frac{p^2}{49})$, which expands to $64 + \frac{32p}{7} + \frac{4p^2}{49} = 80 + \frac{4p^2}{49}$. Solving the resulting linear equation $64 + \frac{32p}{7} = 80$ yields $\frac{32p}{7} = 16$, resulting in $p = \frac{7}{2}$.

The equation of a plane in intercept form that meets the coordinate axes at $A(a,0,0)$, $B(0,b,0)$, and $C(0,0,c)$ is given by $\frac{X}{a} + \frac{Y}{b} + \frac{Z}{c} = 1$. Since this plane passes through the fixed point $(3,2,1)$, the parameters $a, b,$ and $c$ must satisfy the condition $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$. The three planes drawn parallel to the coordinate planes through $A, B,$ and $C$ are defined by the equations $X=a$, $Y=b$, and $Z=c$, respectively. The point of intersection $(x,y,z)$ of these three planes is therefore $(a,b,c)$, meaning $x=a$, $y=b$, and $z=c$. Substituting these coordinates into the established condition, the locus of the intersection point is $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$.

The equation of a plane perpendicular to a line segment is determined by its midpoint and the segment vector acting as the normal vector. With points $(1,2,3)$ and $(-3,4,5)$, the midpoint is calculated as $(-1, 3, 4)$ and the normal vector is $(-4, 2, 2)$, which simplifies to $(2, -1, -1)$ by dividing by $-2$. Substituting these values into the plane equation $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ results in $2(x+1) - (y-3) - (z-4) = 0$, which simplifies to $2x - y - z + 9 = 0$. Inserting the coordinates from option C into this equation results in $2(-3) - 2 - 1 + 9 = -6 - 2 - 1 + 9 = 0$, demonstrating that the point $(-3, 2, 1)$ lies on the plane.

The equation of a plane containing a line and a point is determined by its normal vector, which is the cross product of two vectors lying in the plane. Using the given line points $A(1, -1, 2)$ and $B(1, 1, 1)$, and the external point $P(-2, -2, 2)$, two such vectors are $\vec{AB} = (1-1, 1-(-1), 1-2) = (0, 2, -1)$ and $\vec{AP} = (-2-1, -2-(-1), 2-2) = (-3, -1, 0)$.

The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{AB} \times \vec{AP}$:
$\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -1 \\ -3 & -1 & 0 \end{vmatrix} = \mathbf{i}(0-1) - \mathbf{j}(0-3) + \mathbf{k}(0+6) = -1\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}$
The equation of the plane passing through $A(1, -1, 2)$ with normal vector $(-1, 3, 6)$ is $-1(x-1) + 3(y+1) + 6(z-2) = 0$, which simplifies to $-x + 3y + 6z = 8$. Dividing this equation by 8 yields the intercept form $\frac{x}{-8} + \frac{y}{8/3} + \frac{z}{4/3} = 1$, where the intercepts are $a = -8$, $b = \frac{8}{3}$, and $c = \frac{4}{3}$. The sum of the intercepts is $-8 + \frac{8}{3} + \frac{4}{3} = -8 + \frac{12}{3} = -8 + 4 = -4$.

To determine the length of the projection of a line segment onto a plane, we use the fact that the projection length is given by the product of the segment's length and the sine of the angle between the segment and the plane's normal. First, calculate the vector representing the line segment joining the points $(5, -1, 4)$ and $(4, -1, 3)$, which is $\vec{AB} = (4-5, -1-(-1), 3-4) = (-1, 0, -1)$. The magnitude of this segment is $|\vec{AB}| = \sqrt{(-1)^2 + 0^2 + (-1)^2} = \sqrt{2}$.

The normal vector to the plane $x+y+z=7$ is $\vec{n} = (1, 1, 1)$, which has a magnitude of $|\vec{n}| = \sqrt{1^{2+}1^{2+}1^2} = \sqrt{3}$. We find the cosine of the angle $\theta$ between the segment and the normal using the dot product formula, $\cos\theta = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{AB}| |\vec{n}|} = \frac{|(-1)(1) + (0)(1) + (-1)(1)|}{\sqrt{2} \cdot \sqrt{3}} = \frac{|-2|}{\sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$.

Using the trigonometric identity $\sin\theta = \sqrt{1 - \cos^2\theta}$, we determine $\sin\theta = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$. Finally, the length of the projection is obtained by multiplying the length of the segment by $\sin\theta$, which gives $\sqrt{2} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.

To find the perpendicular distance from a point to a plane, one must first determine the plane's equation by identifying its normal vector, which is the cross product of the direction vectors of the lines to which the plane is parallel. Calculating the cross product of $\langle 1, -2, 3 \rangle$ and $\langle 2, -1, -1 \rangle$ yields the normal vector $\vec{n} = 5\hat{i} + 7\hat{j} + 3\hat{k}$.

Using the point $(1, -1, -1)$ on the plane, the equation is constructed as $5(x-1) + 7(y+1) + 3(z+1) = 0$, which simplifies to $5x + 7y + 3z + 5 = 0$. Applying the point-to-plane distance formula for the point $(1, 3, -7)$, the calculation becomes $\frac{|5(1) + 7(3) + 3(-7) + 5|}{\sqrt{5^2 + 7^2 + 3^2}}$, resulting in a final value of $\frac{10}{\sqrt{83}}$.

The image $Q$ of point $P$ lies along a line passing through $P$ with direction vector $(1, 4, 5)$ such that the midpoint $F$ of segment $PQ$ rests exactly on the given plane $2x+3y-4z+22=0$. By expressing the line through $P(1, -2, 3)$ parametrically as $(\lambda+1, 4\lambda-2, 5\lambda+3)$ and substituting these coordinates into the plane equation, we solve $2(\lambda+1)+3(4\lambda-2)-4(5\lambda+3)+22=0$ to find $\lambda=1$, which places the midpoint $F$ at $(2, 2, 8)$. Calculating the distance $PF$ as $\sqrt{(2-1)^2 + (2-(-2))^2 + (8-3)^2}$ yields $\sqrt{1^2 + 4^2 + 5^2} = \sqrt{42}$, so the total length $PQ$ is $2\sqrt{42}$.