Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

A plane parallel to two given lines must have a normal vector perpendicular to both lines, which is determined by their cross product. Calculating the cross product of the direction vectors $\vec{d}_1 = (-2, 1, -3)$ and $\vec{d}_2 = (-1, 2, -2)$ results in the normal vector $\vec{n} = (4, -1, -3)$. Since the plane contains the point $(2, 2, -2)$, we write the equation of the plane using the point-normal form:
$4(x - 2) - 1(y - 2) - 3(z + 2) = 0$

Simplifying this expression yields $4x - y - 3z = 12$. To find the intercepts with the coordinate axes, we convert this into the intercept form $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$ by dividing the entire equation by 12:
$\frac{x}{3} + \frac{y}{-12} + \frac{z}{-4} = 1$
The intercepts are $\alpha = 3$, $\beta = -12$, and $\gamma = -4$. Their sum is $p = 3 - 12 - 4 = -13$. The volume $V$ of the tetrahedron formed by the origin and these intercepts is calculated as $V = \frac{1}{6} |\alpha \beta \gamma|$, which gives $V = \frac{1}{6} |3 \times (-12) \times (-4)| = 24$. Consequently, the ordered pair $(V, p)$ is $(24, -13)$.

Two lines are coplanar if the vector connecting a point on each line is perpendicular to the normal vector of the plane, a condition satisfied when the scalar triple product of the difference of the lines' position vectors and their respective direction vectors equals zero. Given lines passing through $(1, 2, 3)$ and $(-26, -18, -28)$ with direction vectors $(\lambda, 1, 2)$ and $(-2, 3, \lambda)$, the coplanarity condition is expressed by the following determinant:

$\begin{vmatrix} 27 & 20 & 31 \\ \lambda & 1 & 2 \\ -2 & 3 & \lambda \end{vmatrix} = 0$

Expanding this determinant yields a quadratic equation in $\lambda$, which simplifies to $(\lambda-3)^2 = 0$, confirming that $\lambda = 3$. With the direction vectors identified as $(3, 1, 2)$ and $(-2, 3, 3)$, the normal vector to the plane is found by taking their cross product:

$\vec{n} = (3, 1, 2) \times (-2, 3, 3) = (-3, -13, 11)$

Using the point $(1, 2, 3)$ and the normal vector $(-3, -13, 11)$, the equation of the plane becomes $-3(x-1) - 13(y-2) + 11(z-3) = 0$, which simplifies to $3x + 13y - 11z + 4 = 0$. By substituting the given options into this plane equation, we can determine which point does not lie on the plane. Testing the point $(0, 4, 5)$ results in $3(0) + 13(4) - 11(5) + 4 = 52 - 55 + 4 = 1$, which is not equal to zero, indicating that this point lies outside the plane.

Parametric representation is the most direct method to identify the intersection points of a line with fixed direction ratios and two given lines in three-dimensional space. By expressing the coordinates of any point $A$ on the line $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}$ as $(3\lambda+7, -\lambda+1, \lambda-2)$ and any point $B$ on the line $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}$ as $(2\mu, 3\mu+7, \mu)$, we define the potential intersection locations based on parameters $\lambda$ and $\mu$. The direction ratios of the segment connecting $A$ and $B$ are obtained by subtracting the coordinates of $A$ from $B$, which results in the vector components $(2\mu-3\lambda-7, 3\mu+\lambda+6, \mu-\lambda+2)$.

Because the intersecting line is given as having direction ratios 1, -4, and 2, these components must be proportional to these specific values. Equating these proportions leads to the following system:
$\frac{2\mu-3\lambda-7}{1} = \frac{3\mu+\lambda+6}{-4} = \frac{\mu-\lambda+2}{2}$
Solving this system of equations consistently yields the parameter values $\lambda = -5$ and $\mu = -3$. Substituting these parameters back into the parametric forms reveals the coordinates of the intersection points to be $A(-8, 6, -7)$ and $B(-6, -2, -3)$.

The final step is to calculate the squared distance $(AB)^2$ using the distance formula $(x_B-x_A)^2 + (y_B-y_A)^2 + (z_B-z_A)^2$. Applying the coordinates, this calculation becomes $(-6 - (-8))^2 + (-2 - 6)^2 + (-3 - (-7))^2$, which simplifies to $2^2 + (-8)^2 + 4^2$. Summing these values gives $4 + 64 + 16$, resulting in a final value of 84.

The locus of the foot of the perpendicular from a point on a circle to a plane is defined by the property that the vector from the point to its projection on the plane is parallel to the normal vector of that plane. Any point on the circle $x^{2+}y^2=1$ at $z=0$ can be represented as $(\cos\theta, \sin\theta, 0)$. The coordinates of its projection $(h, k, l)$ onto the plane $2x+3y+z=6$ satisfy the line equation passing through this point in the direction of the normal vector $(2, 3, 1)$, resulting in the proportionality $\frac{h-\cos\theta}{2} = \frac{k-\sin\theta}{3} = \frac{l}{1} = r$.

From this, the coordinates are expressed as $h = \cos\theta + 2r$, $k = \sin\theta + 3r$, and $l = r$, which leads to $\cos\theta = h-2l$ and $\sin\theta = k-3l$. Applying the identity $\cos^2\theta + \sin^2\theta = 1$ gives the condition $(h-2l)^2 + (k-3l)^2 = 1$. Because the projection must also lie on the plane, we substitute $l = 6-2h-3k$ into this equation. By replacing $h$ with $x$ and $k$ with $y$, we obtain $(x-2(6-2x-3y))^2 + (y-3(6-2x-3y))^2 = 1$, which simplifies to $(5x+6y-12)^2 + 4(3x+5y-9)^2 = 1$ subject to the plane constraint $z=6-2x-3y$.

The reflection of a point $(a, b, c)$ across a plane $Ax + By + Cz + D = 0$ results in an image $(x', y', z')$ such that the ratios $\frac{x'-a}{A} = \frac{y'-b}{B} = \frac{z'-c}{C}$ are equivalent to $-2 \frac{Aa + Bb + Cc + D}{A^2 + B^2 + C^2}$. Applying this principle to the plane $3x - 4y + 12z + 19 = 0$ with the point $(a, b, c)$ and its image $(a-6, \beta, \gamma)$, we equate the difference in x-coordinates to the plane's normal component: $\frac{(a-6)-a}{3} = \frac{-6}{3} = -2$.

Because the ratios are constant, we can solve for $\beta$ and $\gamma$ by setting $\frac{\beta-b}{-4} = -2$ and $\frac{\gamma-c}{12} = -2$, which yields $\beta = b + 8$ and $\gamma = c - 24$. Furthermore, the constant ratio is equal to the expression $\frac{-2(3a - 4b + 12c + 19)}{3^2 + (-4)^2 + 12^2}$. Substituting the value $-2$ for this expression and simplifying the denominator $3^2 + (-4)^2 + 12^2 = 169$ leads to $-2 = \frac{-2(3a - 4b + 12c + 19)}{169}$, which simplifies to $3a - 4b + 12c + 19 = 169$, or $3a - 4b + 12c = 150$.

To find the required value, we use the constraint $a + b + c = 5$ by substituting $a = 5 - b - c$ into the equation $3a - 4b + 12c = 150$, resulting in $3(5 - b - c) - 4b + 12c = 150$. This simplifies to $15 - 7b + 9c = 150$, or $7b - 9c = -135$. Evaluating the target expression $7\beta - 9\gamma$ by substituting our expressions for $\beta$ and $\gamma$ gives $7(b + 8) - 9(c - 24)$, which expands to $7b + 56 - 9c + 216$. Combining the constant terms and substituting $-135$ for $7b - 9c$ yields $-135 + 56 + 216$, which results in 137.

Any plane belonging to the family defined by the intersection of $2x+y-5z=0$ and $3x-y+4z-7=0$ takes the following form:
$(2+3\lambda)x + (1-\lambda)y + (-5+4\lambda)z - 7\lambda = 0$
Rotating this plane by $\frac{\pi}{2}$ makes it perpendicular to the original plane $2x+y-5z=0$, which requires the dot product of their normal vectors $(2, 1, -5)$ and $(2+3\lambda, 1-\lambda, -5+4\lambda)$ to be zero. Solving $2(2+3\lambda) + 1(1-\lambda) - 5(-5+4\lambda) = 0$ yields $\lambda = 2$. Substituting $\lambda = 2$ back into the family equation results in $8x - y + 3z - 14 = 0$, which is satisfied by the point $(1, 0, 2)$ since $8(1) - 0 + 3(2) - 14 = 0$.

To identify the foot of the perpendicular from a point to a line formed by the intersection of two planes, we rely on the condition that the vector connecting the external point to any point on the line must be orthogonal to the direction vector of the line itself. First, we determine the direction vector of the line of intersection by taking the cross product of the normal vectors of the two planes, $\vec{n}_1 = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} + \hat{j} - \hat{k}$. This calculation yields $\vec{v} = (-\hat{i} + 5\hat{j} + 3\hat{k})$, which serves as the direction of the line. We find a specific point on the line by setting $z = 0$ in the equations $x - y = 2$ and $2x + y = 2$, which results in the point $Q\left(\frac{4}{3}, -\frac{2}{3}, 0\right)$.

We represent any point $F$ on the line parametrically as $\left(\frac{4}{3} - \lambda, -\frac{2}{3} + 5\lambda, 3\lambda\right)$. The vector $\vec{AF}$ connecting the given point $A(1, 2, 0)$ to this arbitrary point $F$ is $\left(\frac{1}{3} - \lambda, 5\lambda - \frac{8}{3}, 3\lambda\right)$. For $P$ to be the foot of the perpendicular, this vector must be perpendicular to the direction vector $\vec{v}$, satisfying the condition $\vec{AF} \cdot \vec{v} = 0$. Substituting the vectors, we get $-1\left(\frac{1}{3} - \lambda\right) + 5\left(5\lambda - \frac{8}{3}\right) + 3(3\lambda) = 0$, which simplifies to $-\frac{1}{3} + \lambda + 25\lambda - \frac{40}{3} + 9\lambda = 0$, leading to $35\lambda = \frac{41}{3}$ or $\lambda = \frac{41}{105}$.

The coordinates of the foot of the perpendicular $P(\alpha, \beta, \gamma)$ are determined by the parameter $\lambda$, where $\alpha = \frac{4}{3} - \lambda$, $\beta = -\frac{2}{3} + 5\lambda$, and $\gamma = 3\lambda$. Summing these coordinates provides $\alpha + \beta + \gamma = \frac{2}{3} + 7\lambda$. Substituting the value of $\lambda = \frac{41}{105}$, the sum becomes $\frac{2}{3} + 7\left(\frac{41}{105}\right) = \frac{2}{3} + \frac{41}{15} = \frac{10+41}{15} = \frac{51}{15}$. Multiplying this result by $35$ gives $35 \times \frac{51}{15} = 7 \times 17$, which equals $119$.

Determining the distance between a point and the intersection of a line and a plane relies on first identifying the coordinates of the intersection point by parameterizing the line and substituting it into the plane's equation. Representing the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ through a parameter $\lambda$ provides the general point $(\lambda+3, 2\lambda+4, 2\lambda+5)$, which must satisfy the plane equation $x+y+z=17$ at the point of intersection. Substituting these variables into the plane equation yields $(\lambda+3) + (2\lambda+4) + (2\lambda+5) = 17$, which simplifies to $5\lambda + 12 = 17$, or $\lambda = 1$. The specific intersection coordinate is therefore $(4, 6, 7)$, and applying the distance formula between this point and $(1, 1, 9)$ yields $\sqrt{(4-1)^2 + (6-1)^2 + (7-9)^2}$, which results in $\sqrt{3^2 + 5^2 + (-2)^2} = \sqrt{38}$.

The locus of points equidistant from two planes is determined by the condition that the perpendicular distances to each plane are equal, expressed as:
$\frac{x-2y-2z+1}{\sqrt{1^{2+}(-2)^{2+}(-2)^2}} = \pm \frac{2x-3y-6z+1}{\sqrt{2^{2+}(-3)^{2+}(-6)^2}}$
Simplifying the denominators yields $\frac{x-2y-2z+1}{3} = \pm \frac{2x-3y-6z+1}{7}$, which results in the two bisector equations $x - 5y + 4z + 4 = 0$ and $13x - 23y - 32z + 10 = 0$. Among these, the equation $13x - 23y - 32z + 10 = 0$ represents the acute angle bisector $P$. Substituting the coordinates $\left(-2,0,-\frac{1}{2}\right)$ into this equation gives $13(-2) - 23(0) - 32(-\frac{1}{2}) + 10$, which simplifies to $-26 + 16 + 10 = 0$. Since this expression equals zero, the point lies on the plane $P$.

The equation of any plane passing through the line of intersection of two given planes $P_1 = 0$ and $P_2 = 0$ can be represented as $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar constant. Given the planes defined by $x + y + 4z - 16 = 0$ and $-x + y + z - 6 = 0$, we can express the family of planes containing their intersection as $(x + y + 4z - 16) + \lambda(-x + y + z - 6) = 0$.

Because the required plane passes through the specific coordinate $(1, 2, 3)$, substituting these values into the family equation allows us to determine the scalar $\lambda$: $(1 + 2 + 12 - 16) + \lambda(-1 + 2 + 3 - 6) = 0$, which simplifies to $-1 - 2\lambda = 0$, giving $\lambda = -\frac{1}{2}$. Substituting this value back into the expression and multiplying by $2$ to clear the fraction yields the final plane equation $3x + y + 7z - 26 = 0$. Testing the provided options against this equation reveals that while points $(3, 3, 2)$, $(6, -6, 2)$, and $(-8, 8, 6)$ satisfy the equality, the point $(4, 2, 2)$ results in $3(4) + 2 + 7(2) - 26 = 2$, proving it is not located on the plane.

To find the position vector of $A$, first identify the intersection point $T$ by equating the parametric forms of lines $PR$ and $QS$. These lines are represented as $(3+4\lambda, -1-\lambda, 2+2\lambda)$ and $(1-2\mu, 2+\mu, -4-2\mu)$ respectively; setting these equal leads to the system $\lambda=2$ and $\mu=-5$, which uniquely determines $T$ as $(11, -3, 6)$. Because the vector $TA$ is perpendicular to both lines, its direction is determined by the cross product of the direction vectors $(4, -1, 2)$ and $(-2, 1, -2)$, which yields $(0, -4, -2)$. Given that the length of $TA$ is $\sqrt{5}$, we scale this direction vector by $k = \pm \frac{1}{2}$ to find possible coordinates for $A$ as $T + k(0, -4, -2)$, resulting in $(11, -5, 5)$ or $(11, -1, 7)$. Calculating the magnitude of the position vector $A$ in either case gives $\sqrt{11^2 + 5^2 + 5^2}$ or $\sqrt{11^2 + 1^2 + 7^2}$, both of which equal $\sqrt{171}$.

A plane perpendicular to two other planes must have a normal vector that is the cross product of the normal vectors of those two planes. Identifying the normal vectors for the given equations $3x + y - 2z = 5$ and $2x - 5y - z = 7$ as $\vec{n_1} = (3, 1, -2)$ and $\vec{n_2} = (2, -5, -1)$, we compute the cross product $\vec{n} = \vec{n_1} \times \vec{n_2}$ to find the resulting vector components $(-11, -1, -17)$. Using this normal vector with the point $(1, 2, -3)$ in the standard plane equation form $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$, we obtain $-11(x - 1) - 1(y - 2) - 17(z + 3) = 0$. Expanding these terms yields $-11x + 11 - y + 2 - 17z - 51 = 0$, which simplifies to $11x + y + 17z + 38 = 0$.

Any plane parallel to $x-2y+2z-3=0$ maintains the same normal vector components, allowing us to represent the family of such planes as $x-2y+2z+\lambda=0$. Because the distance from the point $(1, 2, 3)$ to any of these planes is specified as $1$, we use the distance formula $\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^{2+}b^{2+}c^2}} = 1$ to isolate $\lambda$. Substituting the coordinates and coefficients gives $\frac{|1 - 2(2) + 2(3) + \lambda|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$, which simplifies to $\frac{|\lambda + 3|}{3} = 1$. Solving the absolute value equation $|\lambda + 3| = 3$ provides two potential values for the constant, $\lambda = 0$ or $\lambda = -6$.

Mapping the plane equation $x-2y+2z+\lambda=0$ to the standard form $ax+by+cz+d=0$ provides $a=1$, $b=-2$, and $c=2$. Substituting $d=-6$ into the condition $(b-d)=K(c-a)$ yields $-2 - (-6) = K(2 - 1)$, which simplifies to $4 = K(1)$, resulting in $K=4$. The alternate case of $d=0$ would yield $K=-2$, but since the problem requires the positive value, $K=4$ is the determined result.

When a line is perpendicular to a plane, any vector contained within that plane must be orthogonal to the direction vector of the line, meaning their dot product is zero. We define a vector lying on the plane by connecting the two given points $(\lambda, 2, 1)$ and $(4, -2, 2)$, which yields the vector $\vec{v}_{plane} = (4 - \lambda, -4, 1)$. The direction vector of the line connecting $(-2, -21, 29)$ and $(-1, -16, 23)$ is determined by the difference of their coordinates, resulting in $\vec{v}_{line} = (1, 5, -6)$.

Because the plane is perpendicular to the line, the dot product $\vec{v}_{plane} \cdot \vec{v}_{line} = 0$ must hold. This provides the equation $(4 - \lambda)(1) + (-4)(5) + (1)(-6) = 0$, which simplifies to $4 - \lambda - 20 - 6 = 0$, leading to $\lambda = -22$. With this value established, we substitute it into the expression $\left(\frac{\lambda}{11}\right)^2 - \frac{4\lambda}{11} - 4$. Since $\frac{\lambda}{11} = -2$, the calculation becomes $(-2)^2 - 4(-2) - 4$, which evaluates to $4 + 8 - 4$ for a final result of $8$.

Calculating the shortest distance between skew lines requires defining the lines completely and identifying their mutual perpendicular direction. The vector extending from the point $(4,3,8)$ to the foot of the perpendicular $(3,5,7)$ has components $(-1, 2, -1)$ and must be orthogonal to the direction vector $(l, 3, 4)$ of the line $L_1$. Taking the dot product of these vectors results in $-l + 6 - 4 = 0$, which identifies the direction ratio $l$ as $2$. Substituting the foot coordinates $(3,5,7)$ back into the parametric equations allows for solving the variables $a=1$ and $b=3$, fully defining the line $L_1$ as $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$.

With the two lines defined, the cross product of their direction vectors $(2, 3, 4)$ and $(3, 4, 5)$ yields the common normal vector

. The shortest distance corresponds to the magnitude of the projection of the vector connecting points $(1, 2, 3)$ and $(2, 4, 5)$ onto this normal direction. This connecting vector is $(1, 2, 2)$, and performing the scalar projection calculation $\frac{|(1, 2, 2) \cdot (-1, 2, -1)|}{\sqrt{(-1)^2 + 2^2 + (-1)^2}}$ results in the final value $\frac{1}{\sqrt{6}}$.

Determining the angle between lines defined by direction cosines requires solving the simultaneous equations while respecting the fundamental property that the sum of the squares of the direction cosines is unity, specifically $l^{2+}m^{2+}n^2=1$. Substituting $n = l+m$ from the first equation into the second equation, $l^{2+}m^{2-}n^2=0$, results in $l^{2+}m^{2-}(l+m)^2 = 0$, which simplifies to $-2lm = 0$. This implies that either $l=0$ or $m=0$.

When $l=0$, the relation $n=l+m$ reduces to $n=m$, and substituting into the identity $l^{2+}m^{2+}n^2=1$ yields $2m^2=1$, meaning $m=n=\pm 1/\sqrt{2}$. This identifies a direction vector $\mathbf{v_1} = (0, 1/\sqrt{2}, 1/\sqrt{2})$. Conversely, when $m=0$, we have $n=l$, which similarly yields $2l^2=1$, so $l=n=\pm 1/\sqrt{2}$, identifying a second direction vector $\mathbf{v_2} = (1/\sqrt{2}, 0, 1/\sqrt{2})$.

The cosine of the angle $\alpha$ between these lines is the dot product of these unit vectors, $\cos \alpha = \mathbf{v_1} \cdot \mathbf{v_2}$, which gives $\cos \alpha = (0)(1/\sqrt{2}) + (1/\sqrt{2})(0) + (1/\sqrt{2})(1/\sqrt{2}) = 1/2$. Consequently, $\cos^2 \alpha = 1/4$ and $\sin^2 \alpha = 1 - 1/4 = 3/4$. Calculating $\sin^4 \alpha + \cos^4 \alpha$ leads to $(3/4)^2 + (1/4)^2 = 9/16 + 1/16 = 10/16$, which simplifies to $5/8$.

Two planes are parallel if the ratios of their respective $x$, $y$, and $z$ coefficients are identical. By comparing the normal vector components of $P_{1}: 3x+15y+21z=9$ and $P_{3}: 2x+10y+14z=5$, we observe that the ratios $\frac{3}{2} = \frac{15}{10} = \frac{21}{14}$ result in a common value of 1.5. This consistent proportionality confirms that the normal vectors for $P_{1}$ and $P_{3}$ point in the same direction, establishing that these two planes are parallel.