Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

The reflection of point $A$ in a line implies that the vector connecting $A$ and its reflection $A'$ is perpendicular to the line and that the midpoint of the segment $AA'$ must lie on the line itself. The given line has direction ratios $(3, 2, 1)$. Subtracting the coordinates of $A(\alpha, 2\alpha, 1)$ from $A'(2\alpha+1, \alpha^{2-}3\alpha, \frac{\alpha-1}{2})$ yields the vector $(\alpha+1, \alpha^{2-}5\alpha, \frac{\alpha-3}{2})$. Imposing the condition that the dot product of this vector and the line direction vector equals zero gives $3(\alpha+1) + 2(\alpha^{2-}5\alpha) + 1(\frac{\alpha-3}{2}) = 0$. Solving this simplifies to $4\alpha^2 - 13\alpha + 3 = 0$, which factors as $(4\alpha-1)(\alpha-3) = 0$, leading to possible values $\alpha = 3$ and $\alpha = \frac{1}{4}$. The midpoint $M$ of $AA'$ is defined as $(\frac{3\alpha+1}{2}, \frac{\alpha^{2-}\alpha}{2}, \frac{\alpha+1}{4})$. For $\alpha = 3$, the midpoint becomes $(5, 3, 1)$, which satisfies the line equation because $\frac{5-2}{3} = 1$, $\frac{3-1}{2} = 1$, and $\frac{1}{1} = 1$. Since $\alpha = \frac{1}{4}$ results in a midpoint that does not lie on the line, the only valid solution is $\alpha = 3$.

The direction vector $\vec{v}$ of line $L$ is the cross product of the direction vectors of the given lines, $\vec{v_1} = \langle 4, 1, 1 \rangle$ and $\vec{v_2} = \langle 1, 1, 0 \rangle$, which yields $\vec{v} = \vec{v_1} \times \vec{v_2} = \langle -1, 1, 3 \rangle$. Line $L$ passing through $P(1, 1, 1)$ is defined by the parametric equations $x = 1 - t$, $y = 1 + t$, $z = 1 + 3t$, which intersects the $yz$-plane ($x=0$) at $t=1$, resulting in the point $Q(0, 2, 4)$. The line parallel to $L$ passing through $S(1, 0, -1)$ is defined by $x = 1 - k$, $y = k$, $z = -1 + 3k$, which intersects the $yz$-plane at $k=1$, resulting in the point $R(0, 1, 2)$.

The vectors representing the sides of the parallelogram $PQRS$ are $\vec{PQ} = Q - P = \langle -1, 1, 3 \rangle$ and $\vec{PS} = S - P = \langle 0, -1, -2 \rangle$. The area of the parallelogram is equal to the magnitude of the cross product of these side vectors,

. The area is $|\langle 1, -2, 1 \rangle| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$, so the square of the area is 6.

The distance $d$ from a point $Q$ to a line passing through point $P$ with direction vector $\vec{b}$ is given by $d = \frac{|\vec{PQ} \times \vec{b}|}{|\vec{b}|}$. Given the line makes equal angles with the coordinate axes, its direction vector is $\vec{b} = (1, 1, 1)$, with magnitude $|\vec{b}| = \sqrt{3}$. Using $P(-3, 5, 2)$ and $Q(-2, r, 1)$, the vector $\vec{PQ}$ is $(1, r-5, -1)$, and the cross product is calculated as:
$\vec{PQ} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & r-5 & -1 \\ 1 & 1 & 1 \end{vmatrix} = (r-4)\mathbf{i} - 2\mathbf{j} + (6-r)\mathbf{k}$

The square of the magnitude of this cross product is $(r-4)^2 + (-2)^2 + (6-r)^2 = 2r^2 - 20r + 56$. Equating the distance $d = \frac{\sqrt{2r^2 - 20r + 56}}{\sqrt{3}}$ to $\sqrt{\frac{14}{3}}$ leads to:
$2r^2 - 20r + 56 = 14$
Simplifying the quadratic equation yields $2r^2 - 20r + 42 = 0$, or $r^2 - 10r + 21 = 0$. By the properties of quadratic equations, the sum of all possible values for $r$ is $-\frac{b}{a} = 10$.

A line perpendicular to two given lines possesses a direction vector defined by the cross product of the direction vectors of those lines. For the lines $L_1$ and $L_2$, the direction vectors are $\langle 3, 5, 7 \rangle$ and $\langle 1, 4, 7 \rangle$, so their cross product is calculated as

, which simplifies to the direction vector $\vec{n} = \langle 1, -2, 1 \rangle$. With $L_3$ having a direction vector $\vec{b}_3 = \langle 2, 1, 2 \rangle$, the angle $\theta$ between the lines satisfies $\cos \theta = \frac{|\vec{n} \cdot \vec{b}_3|}{|\vec{n}| |\vec{b}_3|}$. Inserting the values gives $\frac{|(1)(2) + (-2)(1) + (1)(2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{2^2 + 1^2 + 2^2}} = \frac{2}{3\sqrt{6}}$. Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{4}{54}} = \frac{\sqrt{50}}{\sqrt{54}} = \frac{5\sqrt{2}}{3\sqrt{6}}$, the final value is $\tan \theta = \frac{5\sqrt{2}}{3\sqrt{6}} \div \frac{2}{3\sqrt{6}}$, which simplifies to $\frac{5}{2}\sqrt{2}$.

The parametric coordinates of any point $C$ on line $L_1$ are $(2-3s, 6+2s, 7+4s)$ and any point $D$ on line $L_2$ are $(4+2t, 3+t, 5+3t)$. The vector connecting these points is $\overrightarrow{CD} = (2+2t+3s)\hat{i} + (-3+t-2s)\hat{j} + (-2+3t-4s)\hat{k}$. Since $L_3$ is parallel to the vector $-3\hat{i}+5\hat{j}+16\hat{k}$, the vector $\overrightarrow{CD}$ must satisfy $\overrightarrow{CD} = k(-3\hat{i}+5\hat{j}+16\hat{k})$ for some scalar $k$. Equating the components results in the system of linear equations $2+2t+3s = -3k$, $-3+t-2s = 5k$, and $-2+3t-4s = 16k$, which solves to $k=1$, $s=-3$, and $t=2$. Substituting these values, the points are $C = (11, 0, -5)$ and $D = (8, 5, 11)$, which confirms $\overrightarrow{CD} = -3\hat{i}+5\hat{j}+16\hat{k}$. The squared magnitude is $|\overrightarrow{CD}|^2 = (-3)^2 + 5^2 + 16^2 = 9 + 25 + 256 = 290$.

The distance from the point $P(-2, -8, 6)$ to a line along a specified direction implies that we are constructing a path through $P$ that moves parallel to the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$. This path follows the direction ratios $(1, -1, 2)$, allowing us to represent any point $B$ on this path parametrically as $(\lambda - 2, -\lambda - 8, 2\lambda + 6)$.

Because the point $B$ must also lie on the given line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$, we can describe $B$ using a second parameter $\mu$ as $(\mu + 1, 2\mu + 1, -\mu)$. Equating the corresponding coordinates from these two expressions results in a system of equations that, when solved, yields $\mu = -4$ and $\lambda = -1$, confirming the intersection point is $(-3, -7, 4)$. Calculating the distance between $P(-2, -8, 6)$ and $B(-3, -7, 4)$ gives $\sqrt{(-3 - (-2))^2 + (-7 - (-8))^2 + (4 - 6)^2}$, which simplifies to $\sqrt{(-1)^2 + 1^2 + (-2)^2} = \sqrt{6}$. Squaring this magnitude results in $6$.

To find the image of a point in a line, establish that the vector between the original point and the image is perpendicular to the line's direction, and that the midpoint of this segment lies on the line. For the point $(1, 2, 7)$ and the line defined by $\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$, the line's direction vector is $(1, 1, 2)$. Let the image be $(\alpha, \beta, \gamma)$. The perpendicularity condition gives $(\alpha-1, \beta-2, \gamma-7) \cdot (1, 1, 2) = 0$, which simplifies to $\alpha + \beta + 2\gamma = 17$. Additionally, the midpoint $M(\frac{\alpha+1}{2}, \frac{\beta+2}{2}, \frac{\gamma+7}{2})$ must satisfy the line equation, implying $\frac{\alpha+1}{2} = \frac{\beta}{2} = \frac{\gamma+3}{4}$. Solving these equations yields the image coordinates $(3, 4, 5)$.

Now, apply the distance formula between $(a, 2, 5)$ and $(3, 4, 5)$ and set it equal to $4$. This gives $\sqrt{(a-3)^2 + (2-4)^2 + (5-5)^2} = 4$, which simplifies to $(a-3)^2 + 4 = 16$. Expanding the equation $(a-3)^2 = 12$ results in $a^2 - 6a - 3 = 0$. Based on the properties of quadratic roots, the sum of all possible values of $a$ is $6$.

The line equation $\frac{x}{1} = \frac{1-y}{-2} = \frac{z-2}{3}$ is rewritten in symmetric form as $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda$. Since point $B(4, 9, \alpha)$ lies on this line, $x=4$ implies $\lambda=4$, and the line passes through point $P(0, 1, 2)$ with direction vector $\vec{v} = (1, 2, 3)$. The perpendicular distance $h$ from point $A(1, 6, 3)$ to the line is given by $h = \frac{|\vec{PA} \times \vec{v}|}{|\vec{v}|}$, where $\vec{PA} = (1-0, 6-1, 3-2) = (1, 5, 1)$.

The cross product is computed as

. Substituting the magnitudes, $h = \frac{\sqrt{13^2 + (-2)^2 + (-3)^2}}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{\sqrt{169+4+9}}{\sqrt{14}} = \sqrt{\frac{182}{14}} = \sqrt{13}$. Given the base $BC = 10$, the area is $\frac{1}{2} \times 10 \times \sqrt{13} = 5\sqrt{13}$.

Any point residing on a line is uniquely defined by a single parameter, allowing us to represent point P on the first line as $(2\lambda, 3\lambda, 3\lambda-1)$ and point Q on the second line as $(-\mu-1, \mu+2, 4\mu)$. Since the line segment connecting these points follows direction ratios 1, -1, 2, the coordinate differences must satisfy the relationship:
$\frac{-\mu - 2\lambda - 1}{1} = \frac{\mu - 3\lambda + 2}{-1} = \frac{4\mu - 3\lambda + 1}{2}$
Solving this system provides $\lambda = \frac{1}{5}$ and $\mu = -\frac{8}{15}$, leading to specific coordinates $P = (\frac{2}{5}, \frac{3}{5}, -\frac{2}{5})$ and $Q = (-\frac{7}{15}, \frac{22}{15}, -\frac{32}{15})$. The squared distance $\alpha^2$ between these points is calculated as $\alpha^2 = \frac{1014}{225}$, which confirms that $225 \alpha^2 = 1014$.

Any point on the line $L$ can be expressed as $(2k-1, 3k-1, 6k-3)$ by parameterizing the equations in terms of $k$. The line defined by $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ implies $z=9$ and $\frac{x+1}{2}=\frac{y+1}{3}$, which intersects $L$ at the point $P_0(3, 5, 9)$ when $k=2$. The distance along $L$ from any point $P(k)$ on $L$ to the intersection point $P_0$ is calculated using the distance formula between the two points, which is $\sqrt{((2k-1)-3)^2 + ((3k-1)-5)^2 + ((6k-3)-9)^2} = 7$.

This simplifies to $\sqrt{4(k-2)^2 + 9(k-2)^2 + 36(k-2)^2} = 7$, resulting in the equation $7|k-2|=7$. Solving this yields $|k-2|=1$, which gives the parameter values $k=3$ and $k=1$. Substituting these values into the parametric form for $L$, the corresponding points in $S$ are $P_1(5, 8, 15)$ and $P_2(1, 2, 3)$. The sum of the coordinates for these points is $(5+8+15) + (1+2+3) = 28 + 6 = 34$.

A line that is perpendicular to two specific vectors has a direction defined by their cross product. By calculating the cross product of the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$, we identify the direction vector of the line as $\vec{n} = 2\hat{i} - 3\hat{j} + 2\hat{k}$. Since the line passes through the point $(1, 1, 1)$, its equation in parametric form is represented as $\frac{x-1}{2} = \frac{y-1}{-3} = \frac{z-1}{2} = \lambda$.

Any point $P(a, b, c)$ on this line takes the form $(2\lambda + 1, -3\lambda + 1, 2\lambda + 1)$. Because $P$ serves as the foot of the perpendicular dropped from the origin, the vector $\vec{OP}$ must be perpendicular to the direction vector $\vec{n}$, which means their dot product must be zero. This condition is described by:
$2(2\lambda+1) - 3(-3\lambda+1) + 2(2\lambda+1) = 0$
Upon expanding, we get $4\lambda + 2 + 9\lambda - 3 + 4\lambda + 2 = 0$, which simplifies to $17\lambda + 1 = 0$, resulting in $\lambda = -1/17$. The sum of the coordinates $a + b + c$ is equivalent to $(2\lambda + 1) + (-3\lambda + 1) + (2\lambda + 1) = \lambda + 3$. Evaluating $34(a + b + c)$ yields $34(\frac{-1}{17} + 3)$, which simplifies to $-2 + 102 = 100$.

The line equation is $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = k$, where any point on the line can be represented as $(3k+6, 2k+7, -2k+7)$. Let the midpoint $M$ of $P(1, 2, a)$ and $Q(5, b, c)$ be $(\frac{1+5}{2}, \frac{2+b}{2}, \frac{a+c}{2}) = (3, \frac{2+b}{2}, \frac{a+c}{2})$. Since $M$ must lie on the line, its coordinates must satisfy the line equation at $x=3$, where $3k+6 = 3 \implies k=-1$. Substituting $k=-1$ into the line equations for $y$ and $z$ yields $y = 2(-1)+7 = 5$ and $z = -2(-1)+7 = 9$. Thus, $\frac{2+b}{2} = 5 \implies b=8$ and $\frac{a+c}{2} = 9 \implies a+c = 18$.

The vector $\vec{PQ} = (5-1, b-2, c-a) = (4, 6, c-a)$ is perpendicular to the line with direction vector $\vec{v} = (3, 2, -2)$. The dot product $\vec{PQ} \cdot \vec{v} = 4(3) + 6(2) + (c-a)(-2) = 12 + 12 - 2(c-a) = 0$ simplifies to $24 - 2(c-a) = 0$, giving $c-a = 12$. Solving the system $a+c=18$ and $c-a=12$ yields $2c = 30 \implies c = 15$ and $a = 3$. The sum $a^{2+}b^{2+}c^2 = 3^{2+}8^{2+}15^2 = 9+64+225 = 298$.

To determine the centroid of a triangle when two vertices lie on a line at a specific distance from the third vertex, we utilize the parametric form of the line. By setting the equation of the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ equal to a parameter $\lambda$, any point on the line can be represented as $(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$. Since the distance between these points and $R(1, 2, 3)$ is fixed at 6, we set the square of the distance formula equal to $6^2 = 36$.

Substituting the parametric coordinates into the distance formula gives $(8\lambda - 3 - 1)^2 + (2\lambda + 4 - 2)^2 + (2\lambda - 1 - 3)^2 = 36$, which simplifies to $(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36$. Expanding these squares leads to $64\lambda^2 - 64\lambda + 16 + 4\lambda^2 + 8\lambda + 4 + 4\lambda^2 - 16\lambda + 16 = 36$. Collecting like terms yields $72\lambda^2 - 72\lambda = 0$, which simplifies to $72\lambda(\lambda - 1) = 0$. Solving for $\lambda$, we find the distinct values $\lambda = 0$ and $\lambda = 1$. Applying these to the parametric form, we identify the vertices $P(-3, 4, -1)$ and $Q(5, 6, 1)$.

The centroid $(\alpha, \beta, \gamma)$ is calculated as the arithmetic mean of the coordinates of the three vertices. Averaging the coordinates of $P(-3, 4, -1)$, $Q(5, 6, 1)$, and $R(1, 2, 3)$ gives $\alpha = \frac{-3 + 5 + 1}{3} = 1$, $\beta = \frac{4 + 6 + 2}{3} = 4$, and $\gamma = \frac{-1 + 1 + 3}{3} = 1$. The sum $\alpha + \beta + \gamma$ is therefore $1 + 4 + 1$, which equals 6.

When two lines intersect on the $xy$-plane, the $z$-coordinate of the intersection point must be zero. By letting the intersection point be $(\alpha, \beta, 0)$, we can equate the parametric forms of each line to determine the coordinates in terms of $a$ and $b$. For the first line $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$, substituting $z=0$ gives the parameter $k_1 = \frac{b+1}{7}$, which allows us to express the coordinates as $\alpha = 3(\frac{b+1}{7}) - 1$ and $\beta = 5(\frac{b+1}{7}) - a$. Similarly, for the second line $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$, setting $z=0$ provides the parameter $k_2 = \frac{-2a}{7}$, leading to $\alpha = \frac{-2a}{7} + 2$ and $\beta = \frac{-8a}{7} + b$.

Equating the two expressions for $\alpha$ yields $\frac{3b+3}{7} - 1 = \frac{-2a}{7} + 2$, which simplifies to $2a + 3b = 18$. Likewise, equating the expressions for $\beta$ gives $\frac{5b+5}{7} - a = \frac{-8a}{7} + b$, which simplifies to $a - 2b = -5$. Solving this system of linear equations results in $a = 3$ and $b = 4$, and adding these values together produces 7.

The direction cosines $(l, m, n)$ satisfy $n=4l+m$ from the first equation, and substituting this into the second equation $2mn+10nl+3lm=0$ gives $2m(4l+m)+10l(4l+m)+3lm=0$. Expanding and simplifying leads to $8lm+2m^{2+}40l^{2+}10lm+3lm=40l^{2+}21lm+2m^2=0$. Dividing this by $l^2$ provides the quadratic equation $2(m/l)^{2+}21(m/l)+40=0$, which factors as $(2(m/l)+5)((m/l)+8)=0$.

This yields two sets of direction ratios: for $m/l=-5/2$, we take $l_1=2$ and $m_1=-5$, giving $n_1=4(2)-5=3$, resulting in vector $\mathbf{v}_1=(2, -5, 3)$; for $m/l=-8$, we take $l_2=1$ and $m_2=-8$, giving $n_2=4(1)-8=-4$, resulting in vector $\mathbf{v}_2=(1, -8, -4)$. The cosine of the angle $\theta$ between these lines is determined by the formula $\cos\theta = \frac{|\mathbf{v}_1 \cdot \mathbf{v}_2|}{|\mathbf{v}_1||\mathbf{v}_2|}$, which evaluates to $\frac{|(2)(1) + (-5)(-8) + (3)(-4)|}{\sqrt{2^{2+}(-5)^{2+}3^2}\sqrt{1^{2+}(-8)^{2+}(-4)^2}} = \frac{|2+40-12|}{\sqrt{38}\sqrt{81}} = \frac{30}{9\sqrt{38}} = \frac{10}{3\sqrt{38}}$.

The shortest distance $d$ between two skew lines with points $P_1, P_2$ and direction vectors $\vec{v}_1, \vec{v}_2$ is calculated by $d = \frac{|(\vec{P}_2 - \vec{P}_1) \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}$. Substituting the given points $P_1(-1, 2, 4)$ and $P_2(0, 1, 1)$ along with direction vectors $\vec{v}_1 = \langle \alpha, -1, -\alpha \rangle$ and $\vec{v}_2 = \langle \alpha, 2, 2\alpha \rangle$, the vector difference is $\vec{P}_2 - \vec{P}_1 = \langle 1, -1, -3 \rangle$ and the cross product is $\vec{v}_1 \times \vec{v}_2 = \langle 0, -3\alpha^2, 3\alpha \rangle$. The magnitude of the cross product is $|\vec{v}_1 \times \vec{v}_2| = 3|\alpha|\sqrt{\alpha^{2+}1}$, and the absolute value of the dot product is $|3\alpha^2 - 9\alpha| = 3|\alpha||\alpha - 3|$. Equating the resulting distance $\frac{|\alpha-3|}{\sqrt{\alpha^{2+}1}}$ to $\sqrt{2}$ yields the equation $\frac{(\alpha-3)^2}{\alpha^{2+}1} = 2$. Simplifying this gives $\alpha^2 - 6\alpha + 9 = 2\alpha^2 + 2$, which reduces to the quadratic equation $\alpha^2 + 6\alpha - 7 = 0$. The sum of the roots of this quadratic equation is given by -6.

The shortest distance between two skew lines defined by $\vec{r} = \vec{a_1} + \lambda \vec{d_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{d_2}$ is calculated using the formula $d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$.

To determine the point $P(\alpha, \beta, \gamma)$, represent the line $\frac{x-1}{2} = \frac{y+1}{-3} = z = k$ in parametric form as $(2k+1, -3k-1, k)$. Given the distance constraint of $4\sqrt{14}$ from $(1, -1, 0)$, equate the square of the distance to $(4\sqrt{14})^2 = 224$, which yields $(2k)^2 + (-3k)^2 + k^2 = 14k^2 = 224$, leading to $k = \pm 4$. Evaluating the potential points $(9, -13, 4)$ and $(-7, 11, -4)$ for their distance to the origin shows that $(-7, 11, -4)$ is nearer, identifying $\alpha = -7, \beta = 11$, and $\gamma = -4$.