Three Dimensional Geometry – JEE Main MathematicsPractice Questions & PYQs

Any point on a line defined by a position vector and a direction vector can be expressed parametrically. Starting from the coordinate $P(1, -2, 3)$ and moving in the direction of $Q(5, -4, 7)$, the line's direction vector is determined by the difference $(5-1, -4-(-2), 7-3)$, which results in $(4, -2, 4)$. Therefore, any point on this line takes the form $(1+4t, -2-2t, 3+4t)$ for some real parameter $t$.

The distance from $P$ to a point on this line corresponds to the magnitude of the displacement vector $(4t, -2t, 4t)$, which is calculated as $\sqrt{(4t)^2 + (-2t)^2 + (4t)^2}$. This simplifies to $\sqrt{16t^2 + 4t^2 + 16t^2} = \sqrt{36t^2} = 6|t|$. Equating this to 9 gives $|t| = 1.5$, meaning $t$ is either $1.5$ or $-1.5$. These two values of $t$ lead to the coordinates $(7, -5, 9)$ and $(-5, 1, -3)$ respectively.

Evaluating the sum of the squares of the coordinates for each point reveals which is further from the origin. For $(7, -5, 9)$, the value $\alpha^{2+}\beta^{2+}\gamma^2$ is $7^2 + (-5)^2 + 9^2 = 155$. For $(-5, 1, -3)$, the sum is $(-5)^2 + 1^2 + (-3)^2 = 35$. Because 155 is greater than 35, the point $(7, -5, 9)$ is the one further from the origin, confirming that $\alpha^{2+}\beta^{2+}\gamma^2$ equals 155.

The area of a quadrilateral in three-dimensional space can be determined by taking half the magnitude of the cross product of its diagonal vectors. Utilizing the provided diagonal vectors $\vec{AC} = \hat{i} - \hat{j} - 2 \hat{k}$ and $\vec{BD} = \frac{5}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{2}{3} \hat{k}$, the cross product $\vec{BD} \times \vec{AC}$ is calculated using the determinant, which results in the vector $\frac{8}{3} \hat{i} + \frac{8}{3} \hat{j}$. Determining the magnitude of this resultant vector yields $\sqrt{(\frac{8}{3})^2 + (\frac{8}{3})^2}$, which simplifies to $\frac{8 \sqrt{2}}{3}$. Applying the formula by halving this result gives a final area of $\frac{4 \sqrt{2}}{3}$.

A point $B$ on the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ located at a distance from $A(7, -2, 11)$ along the direction $(2, -3, 6)$ must have a position vector such that $\vec{AB}$ is parallel to $(2, -3, 6)$. Using a parameter $k$ for the first line, point $B$ is expressed as $(k+6, 4, 3k+8)$, making the vector $\vec{AB} = (k+6-7, 4-(-2), 3k+8-11) = (k-1, 6, 3k-3)$. Since this vector must be proportional to $(2, -3, 6)$, the component ratios $\frac{k-1}{2} = \frac{6}{-3} = \frac{3k-3}{6}$ must be equal. Equating these gives $\frac{k-1}{2} = -2$, which leads to $k=-3$. This value of $k$ identifies the coordinates of point $B$ as $(3, 4, -1)$, and the distance is calculated as $\sqrt{(7-3)^2 + (-2-4)^2 + (11-(-1))^2} = \sqrt{4^2 + (-6)^2 + 12^2} = \sqrt{16+36+144} = \sqrt{196} = 14$.

Defining the feet of perpendiculars $Q$ and $R$ as general points on their respective lines, we express $Q$ as $(r, r, 1)$ for the line $x = y, z = 1$ which has the direction vector $(1, 1, 0)$, and $R$ as $(k, -k, -1)$ for the line $x = -y, z = -1$ which has the direction vector $(1, -1, 0)$. The vector $\vec{PQ} = (r - a, r - a, 1 - a)$ must be orthogonal to the direction $(1, 1, 0)$, implying the dot product $(r - a)(1) + (r - a)(1) + (1 - a)(0) = 0$, so $2(r - a) = 0$ means $r = a$ and $Q$ is $(a, a, 1)$. Similarly, the vector $\vec{PR} = (k - a, -k - a, -1 - a)$ must be orthogonal to the direction $(1, -1, 0)$, leading to the dot product equation $(k - a)(1) + (-k - a)(-1) + (-1 - a)(0) = 0$, which simplifies to $k - a + k + a = 0$, or $2k = 0$, confirming $k = 0$ and $R$ is $(0, 0, -1)$. With the condition $\angle QPR = 90^\circ$, the dot product $\vec{PQ} \cdot \vec{PR}$ must equal zero. Substituting $\vec{PQ} = (0, 0, 1 - a)$ and $\vec{PR} = (-a, -a, -1 - a)$, the calculation $(0)(-a) + (0)(-a) + (1 - a)(-1 - a) = 0$ reduces to $-(1 - a^2) = 0$, resulting in $a^2 = 1$. Consequently, $12 a^2$ equates to $12(1) = 12$.

The shortest distance between two skew lines is determined by projecting the vector connecting any two points on the lines onto the common perpendicular vector, which is obtained by taking the cross product of the lines' direction vectors. From the given equations, we identify the position vectors $A(2, 1, 3)$ and $B(2, 3, 5)$, which yield the connecting vector $\vec{AB} = 0\hat{i} + 2\hat{j} + 2\hat{k}$. The direction vectors for the lines are $\vec{p} = \hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{q} = 2\hat{i} + 3\hat{j} + \hat{k}$, so their cross product is $\vec{p} \times \vec{q} = -15\hat{i} + 7\hat{j} + 9\hat{k}$. We then find the magnitude of this cross product as $\sqrt{(-15)^2 + 7^2 + 9^2} = \sqrt{355}$. The shortest distance is the absolute value of the dot product $\vec{AB} \cdot (\vec{p} \times \vec{q})$ divided by this magnitude, resulting in $\frac{|0(-15) + 2(7) + 2(9)|}{\sqrt{355}} = \frac{32}{\sqrt{355}}$. With $m=32$ and $n=355$, the sum $m+n$ equals $387$.

Finding the intersection of two lines requires expressing both lines in parametric form and equating their coordinates to identify the common point. The first line $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ simplifies to the symmetric form $\frac{x-2}{1}=\frac{y}{-1}=\frac{z-7}{8}=\lambda$, which yields general coordinates $(2+\lambda, -\lambda, 7+8\lambda)$, while the second line $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}=k$ provides coordinates $(4k-3, 3k-2, k-2)$. Equating these coordinates leads to the system $\lambda+5=4k$ and $-\lambda=3k-2$, which simplifies to $7k=7$, yielding $k=1$ and $\lambda=-1$. Substituting these parameters confirms the intersection point $P$ is $(1, 1, -1)$.

Calculating the distance $l$ from $P$ to the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ involves using a point $A(-1, 1, 1)$ on the line to define the vector $\vec{AP} = (1-(-1), 1-1, -1-1) = (2, 0, -2)$. The squared distance $l^2$ is the difference between the squared magnitude $|\vec{AP}|^2$ and the square of the projection of $\vec{AP}$ onto the line's direction vector $\vec{v}=(2, 3, 1)$. With $|\vec{AP}|^2 = 2^{2+}0^{2+}(-2)^2 = 8$ and the projection magnitude given by $\frac{\vec{AP} \cdot \vec{v}}{|\vec{v}|} = \frac{2(2)+0(3)-2(1)}{\sqrt{2^{2+}3^{2+}1^2}} = \frac{2}{\sqrt{14}}$, the squared distance becomes $l^2 = 8 - \frac{4}{14}$. Consequently, $14l^2 = 8(14) - 4 = 112 - 4 = 108$.

When two lines in space are parallel, the shortest distance between them is equivalent to the perpendicular distance from any point on the first line to the second line. The given lines can be expressed in vector form as $\vec{r}_1 = (\lambda \hat{i} + 4\hat{j} + 3\hat{k}) + \alpha(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r}_2 = (2\hat{i} + 4\hat{j} + 7\hat{k}) + \beta(4\hat{i} + 6\hat{j} + 8\hat{k})$. Since the direction vectors are proportional, with the second being exactly twice the first, they share a common direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

To calculate the distance, we employ the formula $d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}$, where $\vec{a}_1 = \lambda \hat{i} + 4\hat{j} + 3\hat{k}$ and $\vec{a}_2 = 2 \hat{i} + 4\hat{j} + 7\hat{k}$ are position vectors for points on the lines. The difference vector is $\vec{a}_2 - \vec{a}_1 = (2-\lambda)\hat{i} + 0\hat{j} + 4\hat{k}$. The cross product of the common direction vector and this difference vector is calculated as:
$\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 2-\lambda & 0 & 4 \end{vmatrix} = 12\hat{i} - 4\lambda\hat{j} + (3\lambda-6)\hat{k}$

Given the distance $d = \frac{13}{\sqrt{29}}$ and knowing the magnitude $|\vec{b}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$, we equate the magnitude of the cross product divided by $\sqrt{29}$ to the given distance. This yields the equation $\sqrt{12^2 + (-4\lambda)^2 + (3\lambda-6)^2} = 13$. Squaring both sides produces $144 + 16\lambda^2 + (3\lambda-6)^2 = 169$, which simplifies to $16\lambda^2 + (3\lambda-6)^2 = 25$. Expanding the square results in $25\lambda^2 - 36\lambda + 11 = 0$, which factors into $(\lambda - 1)(25\lambda - 11) = 0$, confirming that $\lambda = 1$ is a valid solution.

The mirror image of a point with respect to a line requires finding the foot of the perpendicular, $N$, which acts as the midpoint between the original point $P(3, 4, 9)$ and its image $(\alpha, \beta, \gamma)$. Any point on the given line can be expressed using a parameter $\lambda$ as $(3\lambda + 1, 2\lambda - 1, \lambda + 2)$. Because the vector $\vec{PN}$ from $P(3, 4, 9)$ to $N$ must be perpendicular to the line's direction vector $(3, 2, 1)$, their dot product must be zero. Solving the equation $3(3\lambda - 2) + 2(2\lambda - 5) + 1(\lambda - 7) = 0$ gives $\lambda = \frac{23}{14}$, which locates $N$ at $(\frac{83}{14}, \frac{32}{14}, \frac{51}{14})$.

Using the midpoint formula for $P$ and its image, the relations $\frac{\alpha + 3}{2} = \frac{83}{14}$, $\frac{\beta + 4}{2} = \frac{32}{14}$, and $\frac{\gamma + 9}{2} = \frac{51}{14}$ yield the image coordinates $\alpha = \frac{62}{7}$, $\beta = \frac{4}{7}$, and $\gamma = -\frac{12}{7}$. Summing these values gives $\alpha + \beta + \gamma = \frac{62+4-12}{7} = \frac{54}{7}$. The requested calculation $14(\alpha + \beta + \gamma)$ results in $14 \times \frac{54}{7}$, which simplifies to 108.

The coordinates of points $P$ and $Q$ on the respective lines $x = y+2 = z$ and $x+2 = 2y = 2z$ can be represented as $P(t, t-2, t)$ and $Q(2s-2, s, s)$. Since the line segment $PQ$ possesses fixed direction ratios $2, 1, 2$, the vector components connecting these points must maintain this proportionality, which allows for the equation
$\frac{2s-t-2}{2} = \frac{s-t+2}{1} = \frac{s-t}{2}$
Solving this system of equations yields $t=6$ and $s=2$, which fixes the points at $P(6, 4, 6)$ and $Q(2, 2, 2)$.

The line $PQ$ is then defined by the symmetric equation $\frac{x-2}{2} = \frac{y-2}{1} = \frac{z-2}{2} = \lambda$, so any point $F$ on this line takes the form $(2\lambda+2, \lambda+2, 2\lambda+2)$. We find the perpendicular distance $l$ from the point $A(1, 2, 12)$ to the line by identifying the foot of the perpendicular, $F$, where the vector $\vec{AF}$ is orthogonal to the direction vector $(2, 1, 2)$. This condition satisfies $(2\lambda+1, \lambda, 2\lambda-10) \cdot (2, 1, 2) = 0$, leading to $\lambda=2$ and identifying $F$ as $(6, 4, 6)$. Consequently, the square of the perpendicular distance is calculated as $l^2 = (6-1)^2 + (4-2)^2 + (6-12)^2 = 25+4+36 = 65$.

The fundamental property governing the reflection of a point in a line is that the vector connecting the point and its mirror image must be perpendicular to the direction vector of that line. Considering the given point $P = (2, 3, 5)$ and its mirror image $R = (\alpha, \beta, \gamma)$, the vector $\vec{PR} = (\alpha-2, \beta-3, \gamma-5)$ must be orthogonal to the line's direction vector $\vec{v} = (2, 3, 4)$.

Since perpendicular vectors have a dot product of zero, we set up the equation $\vec{PR} \cdot \vec{v} = 0$, which yields $2(\alpha-2) + 3(\beta-3) + 4(\gamma-5) = 0$. Expanding these terms gives $2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0$. By grouping the variables on one side, we arrive at $2\alpha + 3\beta + 4\gamma = 4 + 9 + 20$, which simplifies to 33.

The mirror image of a point in a line is uniquely defined by the condition that the midpoint of the segment connecting the point and its image must lie on the line, while this segment must also be perpendicular to the line itself. The line $L$ passes through $P(1, 2, 1)$ and $Q(2, 1,-1)$, which provides a direction vector proportional to $P-Q = (-1, 1, 2)$. Using this direction, any point on the line can be represented parametrically as $(1-k, 2+k, 1+2k)$ for some scalar $k$.

The mirror image $A'(\alpha, \beta, \gamma)$ of $A(2, 2, 2)$ implies that the midpoint $M$ of the segment $AA'$ is given by $(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2})$. Since $M$ must lie on line $L$, equating these coordinates to $(1-k, 2+k, 1+2k)$ yields $\alpha = -2k$, $\beta = 2+2k$, and $\gamma = 4k$. Given that $AA'$ is perpendicular to the line, the dot product of the vector $\vec{AA'} = (\alpha-2, \beta-2, \gamma-2)$ and the direction vector $(-1, 1, 2)$ must be zero. Substituting the expressions for $\alpha, \beta,$ and $\gamma$ into the dot product gives $(-2k-2)(-1) + (2k+2-2)(1) + (4k-2)(2) = 0$, which simplifies to $2k+2 + 2k + 8k-4 = 0$, leading to $12k=2$, or $k=1/6$. Evaluating the expression $\alpha+\beta+6\gamma$ by substituting these parametric forms results in $-2k + (2+2k) + 6(4k) = 2+24k$, which equals $2+24(1/6) = 6$.

Finding the distance from a point to a line along a specific path requires locating the point on the line where that path intersects it. The line $L$ passes through $(1, 2, 3)$ and $(2, 3, 5)$, so its direction ratios are $\langle 2-1, 3-2, 5-3 \rangle = \langle 1, 1, 2 \rangle$. Any arbitrary point $B$ on this line can therefore be represented in parametric form as $(1+\lambda, 2+\lambda, 3+2\lambda)$ for some scalar $\lambda$.

The distance is measured along a direction defined by the line $\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$, which has direction ratios $\langle 2, 1, 2 \rangle$. The vector $\vec{PB}$ connecting the given point $P\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ to point $B$ on line $L$ must be parallel to this direction. Constructing $\vec{PB}$ yields $\langle (1+\lambda) - \frac{11}{3}, (2+\lambda) - \frac{11}{3}, (3+2\lambda) - \frac{19}{3} \rangle$, which simplifies to $\left\langle \frac{3\lambda-8}{3}, \frac{3\lambda-5}{3}, \frac{6\lambda-10}{3} \right\rangle$. Because this vector must be proportional to $\langle 2, 1, 2 \rangle$, we can equate the ratios of the components, specifically $\frac{3\lambda-8}{2} = \frac{3\lambda-5}{1}$, which results in $3\lambda - 8 = 6\lambda - 10$ and gives $\lambda = \frac{2}{3}$.

Using $\lambda = \frac{2}{3}$, the intersection point $B$ is calculated as $\left(1+\frac{2}{3}, 2+\frac{2}{3}, 3+\frac{4}{3}\right) = \left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right)$. The distance is then the magnitude of the vector $\vec{PB}$, given by $\sqrt{\left(\frac{5}{3} - \frac{11}{3}\right)^2 + \left(\frac{8}{3} - \frac{11}{3}\right)^2 + \left(\frac{13}{3} - \frac{19}{3}\right)^2} = \sqrt{(-2)^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.

The image $P(\alpha, \beta, \gamma)$ of a point $Q(1, 6, 4)$ in a line is its reflection across the line, where the foot of the perpendicular from $Q$ to the line is the midpoint of the segment connecting $Q$ and $P$. We can define any point $A$ on the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ using a parameter $t$ as $(t, 2t+1, 3t+2)$. By defining the vector $\overrightarrow{QA} = (t-1, 2t-5, 3t-2)$, we impose the condition that this vector must be perpendicular to the line's direction vector $(1, 2, 3)$.

This orthogonality requires their dot product to be zero, specifically $(t-1) + 2(2t-5) + 3(3t-2) = 0$. Solving this equation yields $14t = 17$, so $t = \frac{17}{14}$. Using the midpoint formula where the foot of the perpendicular $A$ is $\frac{P+Q}{2}$, we obtain $P = 2A - Q$, which results in $\alpha = \frac{20}{14}$, $\beta = \frac{12}{14}$, and $\gamma = \frac{102}{14}$. Calculating the final expression gives $2\alpha+\beta+\gamma = \frac{40+12+102}{14} = \frac{154}{14} = 11$.

The angle at vertex $P$ in triangle $PQR$ is determined by treating $PQ$ and $PR$ as vectors and using the dot product formula, where the cosine of the angle is the dot product of the vectors divided by the product of their magnitudes. The vectors originating from $P$ are $\vec{PQ} = (4-3, 6-2, 2-3) = (1, 4, -1)$ and $\vec{PR} = (7-3, 3-2, 2-3) = (4, 1, -1)$. Their dot product is $(1)(4) + (4)(1) + (-1)(-1) = 9$, and the magnitude of each vector is $\sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{18}$ and $\sqrt{4^2 + 1^2 + (-1)^2} = \sqrt{18}$. Dividing the dot product by the product of these magnitudes yields $\cos \theta = \frac{9}{\sqrt{18} \cdot \sqrt{18}} = \frac{9}{18} = \frac{1}{2}$, establishing the angle $\theta$ as $\frac{\pi}{3}$.

The foot of the perpendicular from an external point to a given line corresponds to the unique location on that line where the segment joining the two points is orthogonal to the line's direction vector. By expressing the coordinates of any point on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$ in terms of a parameter $k$, we obtain the general point $(5k-3, 2k+1, 3k-4)$.

The vector connecting the point $(1, 2, 3)$ to this generic point is $(5k-4, 2k-1, 3k-7)$. Because this segment must be perpendicular to the line with direction ratios $(5, 2, 3)$, the dot product of these vectors must be zero, leading to the equation $5(5k-4) + 2(2k-1) + 3(3k-7) = 0$. Solving this results in $25k - 20 + 4k - 2 + 9k - 21 = 0$, which simplifies to $38k = 43$, or $k = \frac{43}{38}$. The sum of the coordinates $\alpha + \beta + \gamma$ is $(5k-3) + (2k+1) + (3k-4) = 10k - 6$. Substituting $k$ into the expression $19(10k - 6)$ results in $190(\frac{43}{38}) - 114$, which simplifies to $5(43) - 114$, yielding a final value of 101.

The shortest distance between two skew lines is determined using the formula
$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}$
where $\vec{a}_1$ and $\vec{a}_2$ are points on the respective lines and $\vec{v}_1$ and $\vec{v}_2$ are their direction vectors. For the given lines $\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$, the points are $P_1(\lambda, 2, 1)$ and $P_2(\sqrt{3}, 1, 2)$, while the direction vectors are $\vec{v}_1 = -2\hat{i} + \hat{j} + \hat{k}$ and $\vec{v}_2 = \hat{i} - 2\hat{j} + \hat{k}$.

Calculating the cross product yields $\vec{v}_1 \times \vec{v}_2 = 3\hat{i} + 3\hat{j} + 3\hat{k}$, which has a magnitude of $|\vec{v}_1 \times \vec{v}_2| = \sqrt{3^2 + 3^2 + 3^2} = 3\sqrt{3}$. The vector connecting the two points is $\vec{P_1P_2} = (\sqrt{3}-\lambda)\hat{i} - \hat{j} + \hat{k}$. Computing the absolute value of the scalar triple product $\vec{P_1P_2} \cdot (\vec{v}_1 \times \vec{v}_2)$ results in $|3(\sqrt{3}-\lambda) - 3 + 3| = |3(\sqrt{3}-\lambda)|$. Equating the distance to 1 gives $1 = \frac{|3(\sqrt{3}-\lambda)|}{3\sqrt{3}}$, which simplifies to $|\sqrt{3}-\lambda| = \sqrt{3}$.

Solving this absolute value equation produces two possibilities: $\sqrt{3}-\lambda = \sqrt{3}$ leads to $\lambda = 0$, and $\sqrt{3}-\lambda = -\sqrt{3}$ leads to $\lambda = 2\sqrt{3}$. Summing these values, the total is $2\sqrt{3}$.

The line passing through $A(4,-6,-2)$ and $B(16,-2,4)$ has a direction vector determined by the difference in coordinates, which is $(12, 4, 6)$. Calculating the magnitude of this vector gives $\sqrt{12^{2+}4^{2+}6^2} = 14$, which allows for the identification of the unit vector as $(\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$.

Positioning a point $P$ at a distance of 21 units from $A$ along this vector requires adding 21 times the unit vector to the coordinates of $A$, resulting in $(4 + 21 \cdot \frac{6}{7}, -6 + 21 \cdot \frac{2}{7}, -2 + 21 \cdot \frac{3}{7})$. Simplifying these expressions yields the point $(22, 0, 7)$.

The distance between $P(22, 0, 7)$ and $Q(4, -12, 3)$ is determined using the distance formula $\sqrt{(22-4)^2 + (0-(-12))^2 + (7-3)^2}$. Computing the squares results in $\sqrt{18^2 + 12^2 + 4^2}$, which simplifies to $\sqrt{324 + 144 + 16} = \sqrt{484}$, ultimately evaluating to 22.