Let the line passing through the points (-1, 2, 1)$(-1, 2, 1)$ and \parallel to the line fracx - 12 = fracy + 13 = fracz4$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line fracx + 23 = fracy - 32 = fracz - 41$\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point P$P$. Then the distance of P$P$ from the point Q(4, -5, 1)$Q(4, -5, 1)$ is :
A.5$5$
B.10$10$
C.5sqrt6$5\sqrt{6}$
D.5sqrt5$5\sqrt{5}$
Solution & Explanation
### Related Formula
The line passing through veca$\vec{a}$ and \parallel to direction vector vecv = lhatmathbfi + mhatmathbfj + nhatmathbfk$\vec{v} = l\hat{\mathbf{i}} + m\hat{\mathbf{j}} + n\hat{\mathbf{k}}$ is written in symmetric form as:
fracx - x_al = fracy - y_am = fracz - z_an$\frac{x - x_a}{l} = \frac{y - y_a}{m} = \frac{z - z_a}{n}$
### Core Logic
Formulate the equation of the line passing through (-1, 2, 1)$(-1, 2, 1)$ with direction vector components (2, 3, 4)$(2, 3, 4)$:
L_1: fracx + 12 = fracy - 23 = fracz - 14 = lambda quad dots (1)$L_1: \frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 1}{4} = \lambda \quad \dots (1)$Intersection of Lines in 3D Space
Any generic point on this line can be written as:
P = (2lambda - 1, \, 3lambda + 2, \, 4lambda + 1)$P = (2\lambda - 1, \, 3\lambda + 2, \, 4\lambda + 1)$
The second given line equation is:
L_2: fracx + 23 = fracy - 32 = fracz - 41 = mu quad dots (2)$L_2: \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} = \mu \quad \dots (2)$
Any generic point on line L_2$L_2$ is:
P' = (3mu - 2, \, 2mu + 3, \, mu + 4)$P' = (3\mu - 2, \, 2\mu + 3, \, \mu + 4)$
### Step 1: Compute the Point of Intersection
At the point of intersection P$P$, equate the coordinates from both lines:
2lambda - 1 = 3mu - 2 implies 2lambda - 3mu = -1 quad dots (3)$2\lambda - 1 = 3\mu - 2 \implies 2\lambda - 3\mu = -1 \quad \dots (3)$3lambda + 2 = 2mu + 3 implies 3lambda - 2mu = 1 quad dots (4)$3\lambda + 2 = 2\mu + 3 \implies 3\lambda - 2\mu = 1 \quad \dots (4)$4lambda + 1 = mu + 4 implies 4lambda - mu = 3 quad dots (5)$4\lambda + 1 = \mu + 4 \implies 4\lambda - \mu = 3 \quad \dots (5)$
Solving equations (3) and (4) simultaneously gives:
lambda = 1, quad mu = 1$\lambda = 1, \quad \mu = 1$
Verify these values using equation (5):
4(1) - 1 = 3 quad (textSatisfied)$4(1) - 1 = 3 \quad (\text{Satisfied})$
Thus, substituting lambda = 1$\lambda = 1$ gives the coordinates of point P$P$:
P = (2(1) - 1, \, 3(1) + 2, \, 4(1) + 1) = (1, 5, 5)$P = (2(1) - 1, \, 3(1) + 2, \, 4(1) + 1) = (1, 5, 5)$
### Step 2: Calculate Euclidean Distance PQ
Find the distance between P(1, 5, 5)$P(1, 5, 5)$ and Q(4, -5, 1)$Q(4, -5, 1)$ using the 3D distance formula:
PQ = sqrt(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2$PQ = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2}$PQ = sqrt3^2 + (-10)^2 + (-4)^2 = sqrt9 + 100 + 16$PQ = \sqrt{3^2 + (-10)^2 + (-4)^2} = \sqrt{9 + 100 + 16}$PQ = sqrt125 = 5sqrt5$PQ = \sqrt{125} = 5\sqrt{5}$
### Pattern Recognition
When finding the intersection point of two 3D lines, always use the third coordinate equation to verify the parameter values obtained from the first two equations to ensure consistency.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Keywords:#intersection point 3D lines#parametric line equations geometry#JEE Main 2025 Morning Q68#Euclidean distance 3D coordinates
More Three Dimensional Geometry Previous-Year Questions — Page 4
Q522025Intersection of Lines
Let a line passing through the point (4,1,0)$(4,1,0)$ [cite: 520] intersect the line L_1:fracx-12=fracy-23=fracz-34$L_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point A(alpha, beta, gamma)$A(\alpha, \beta, \gamma)$ [cite: 522, 523, 524] and the line L_2:x-6=y=-z+4$L_{2}:x-6=y=-z+4$ at the point B(a, b, c)$B(a, b, c)$[cite: 525]. Then the value of the determinant beginvmatrix 1 & 0 & 1 \\ alpha & beta & gamma \\ a & b & c endvmatrix$\begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix}$ is equal to[cite: 527]:
A. 8
B. 16
C. 12
D. 4
Solution
### Related Formula
For three points P$P$, A$A$, and B$B$ to be collinear, their direction vectors must be proportional:
vecPA parallel vecPB implies fracx_A - x_Px_B - x_P = fracy_A - y_Py_B - y_P = fracz_A - z_Pz_B - z_P$\vec{PA} \parallel \vec{PB} \implies \frac{x_A - x_P}{x_B - x_P} = \frac{y_A - y_P}{y_B - y_P} = \frac{z_A - z_P}{z_B - z_P}$Intersection of Lines diagram for Q52 - JEE Main 2025 Morning
### Core Logic
Express general coordinates for A$A$ on L_1$L_1$ and B$B$ on L_2$L_2$ [cite: 1204]:
L_1: fracx-12=fracy-23=fracz-34=p implies A(2p+1, 3p+2, 4p+3)$L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p \implies A(2p+1, 3p+2, 4p+3)$ [cite: 1204, 1206]
L_2: fracx-61=fracy1=fracz-4-1=q implies B(q+6, q, 4-q)$L_2: \frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q \implies B(q+6, q, 4-q)$ [cite: 1204, 1207]
Direction ratios (D.R.) from P(4,1,0)$P(4,1,0)$ [cite: 1208, 1209]:
textD.R. of PA = (2p-3, 3p+1, 4p+3)$\text{D.R. of } PA = (2p-3, 3p+1, 4p+3)$ [cite: 1208]
textD.R. of PB = (q+2, q-1, 4-q)$\text{D.R. of } PB = (q+2, q-1, 4-q)$ [cite: 1209]
Since P, A, B$P, A, B$ lie on the same line [cite: 1210]:
frac2p-3q+2 = frac3p+1q-1 = frac4p+34-q$\frac{2p-3}{q+2} = \frac{3p+1}{q-1} = \frac{4p+3}{4-q}$ [cite: 1210]
### Step 1: Solving the system of equations
Equating expressions to solve for parameters p$p$ and q$q$ [cite: 1211, 1212]:
From first two [cite: 1211]:
2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 implies pq + 8p + 4q - 1 = 0$2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 \implies pq + 8p + 4q - 1 = 0$ [cite: 1211, 1212]
From second and third components [cite: 1212]:
12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 implies 7pq - 16p + 4q - 7 = 0$12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 \implies 7pq - 16p + 4q - 7 = 0$ [cite: 1212]
Solving equations simultaneously gives [cite: 1234]:
p = -1, quad q = 3$p = -1, \quad q = 3$ [cite: 1234]
Substituting parameters back yields the points [cite: 1236]:
A(-1, -1, -1), quad B(9, 3, 1)$A(-1, -1, -1), \quad B(9, 3, 1)$ [cite: 1236]
### Step 2: Evaluating the Determinant
Substitute values of A$A$ and B$B$ into the matrix grid [cite: 1244]:
beginvmatrix 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 endvmatrix$\begin{vmatrix} 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 \end{vmatrix}$ [cite: 1244]
Applying row/column operations (C_3 rightarrow C_3 - C_1$C_3 \rightarrow C_3 - C_1$) [cite: 1244]:
beginvmatrix 1 & 0 & 0 \\ -1 & -1 & 0 \\ 9 & 3 & -8 endvmatrix = 1((-1)(-8) - 0) = 8$\begin{vmatrix} 1 & 0 & 0 \\ -1 & -1 & 0 \\ 9 & 3 & -8 \end{vmatrix} = 1((-1)(-8) - 0) = 8$ [cite: 1244]
### Pattern Recognition
Collinearity problem involving parameter matches across skew line orientations. Reducing fractions systematically prevents higher-degree calculation mistakes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 12 Mathematics: Determinants
Q702025Shortest Distance between Skew Lines
Line L_1$L_{1}$ passes through the point (1, 2, 3)$(1, 2, 3)$ and is parallel to z-axis[cite: 685]. Line L_2$L_{2}$ passes through the point (lambda, 5, 6)$(lambda, 5, 6)$ and is parallel to y-axis[cite: 686]. Let for lambda = lambda_1, lambda_2$\lambda = \lambda_{1}, \lambda_{2}$, lambda_2 < lambda_1$\lambda_{2} < \lambda_{1}$ the shortest distance between the two lines be 3[cite: 698]. Then the square of the distance of the point (lambda_1, lambda_2, 7)$(lambda_{1}, \lambda_{2}, 7)$ from the line L_1$L_{1}$ is[cite: 698, 700]:
A. 40
B. 32
C. 25
D. 37
Solution
### Related Formula
Shortest distance between perpendicular axes vectors:
textS.D. = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$\text{S.D.} = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
### Core Logic
Represent equations of straight lines symmetrically using vector directions [cite: 1418]:
L_1: fracx-10 = fracy-20 = fracz-31$L_1: \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$ [cite: 1418]
L_2: fracx-lambda0 = fracy-51 = fracz-60$L_2: \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$ [cite: 1418]
Evaluating the standard shortest distance configuration formula[cite: 1419, 1420]:
textS.D. = |lambda - 1| = 3 implies lambda - 1 = pm 3$\text{S.D.} = |\lambda - 1| = 3 \implies \lambda - 1 = \pm 3$ [cite: 1420]
lambda = 4 quad textor quad lambda = -2$\lambda = 4 \quad \text{or} \quad \lambda = -2$ [cite: 1420]
Given the condition lambda_2 < lambda_1$\lambda_2 < \lambda_1$ [cite: 698]:
lambda_1 = 4, quad lambda_2 = -2$\lambda_1 = 4, \quad \lambda_2 = -2$ [cite: 1421, 1422]
### Step 1: Distance calculation from line
We need to find the square of distance from point P(4, -2, 7)$P(4, -2, 7)$ to line L_1$L_1$ [cite: 1424].
Any general matching point coordinates tracking along path L_1$L_1$ look like Q(1, 2, t+3)$Q(1, 2, t+3)$ [cite: 1424].
Form a perpendicular projection vector condition [cite: 1425]:
vecPQ = (-3, 4, t-4)$\vec{PQ} = (-3, 4, t-4)$ [cite: 1425]
Since vecPQ cdot hatk = 0 implies t-4 = 0 implies t=4$\vec{PQ} \cdot \hat{k} = 0 \implies t-4 = 0 \implies t=4$ [cite: 1425, 1426].
Thus, the foot of perpendicular is Q(1, 2, 7)$Q(1, 2, 7)$ [cite: 1427].
Evaluate the squared distance component magnitude [cite: 1427]:
PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0 = 9 + 16 = 25$PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0 = 9 + 16 = 25$ [cite: 1427, 1428]
### Pattern Recognition
For lines parallel directly to independent Cartesian coordinate grid lines, the shortest paths are simply direct plane projections.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q562025Shortest Distance Between Two Lines
Let the values of p$p$, for which the shortest distance between the lines fracx + 13 = fracy4 = fracz5$\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$ and vecr = (phati + 2hatj + hatk) + lambda (2hati + 3hatj + 4hatk)$\vec{r} = (p\hat{i} + 2\hat{j} + hat{k}) + \lambda (2hat{i} + 3hat{j} + 4hat{k})$ is frac1sqrt6$\frac{1}{sqrt{6}}$, be a, b, (a < b$a < b$). Then the length of the latus rectum of the ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:
A.9$9$
B.frac32$\frac{3}{2}$
C.frac23$\frac{2}{3}$
D.18$18$
Solution
### Related Formula
The shortest distance between two lines vecr = veca_1 + lambda vecb_1$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and vecr = veca_2 + mu vecb_2$\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by:
d = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
### Core Logic
From the given line equations:
Line 1 passes through veca_1 = -hati + 0hatj + 0hatk$\vec{a}_1 = -\hat{i} + 0\hat{j} + 0\hat{k}$ along vector vecb_1 = 3hati + 4hatj + 5hatk$\vec{b}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
Line 2 passes through veca_2 = phati + 2hatj + hatk$\vec{a}_2 = p\hat{i} + 2\hat{j} + \hat{k}$ along vector vecb_2 = 2hati + 3hatj + 4hatk$\vec{b}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Vector difference:
veca_2 - veca_1 = (p + 1)hati + 2hatj + hatk$\vec{a}_2 - \vec{a}_1 = (p + 1)\hat{i} + 2\hat{j} + \hat{k}$
Computing the cross product vecb_1 times vecb_2$\vec{b}_1 \times \vec{b}_2$:
vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 3 & 4 & 5 \\ 2 & 3 & 4 endvmatrix = hati(16-15) - hatj(12-10) + hatk(9-8) = hati - 2hatj + hatk$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8) = \hat{i} - 2\hat{j} + \hat{k}$
Magnitude |vecb_1 times vecb_2| = sqrt1^2 + (-2)^2 + 1^2 = sqrt6$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.
### Step 1: Applying the Shortest Distance Value
Substitute these into the distance equation:
d = frac|big((p + 1)hati + 2hatj + hatkbig) cdot big(hati - 2hatj + hatkbig)|sqrt6 = frac1sqrt6$d = \frac{|\big((p + 1)\hat{i} + 2\hat{j} + \hat{k}\big) \cdot \big(\hat{i} - 2\hat{j} + \hat{k}\big)|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$|(p + 1)(1) + 2(-2) + 1(1)| = 1$|(p + 1)(1) + 2(-2) + 1(1)| = 1$|p + 1 - 4 + 1| = 1 implies |p - 2| = 1$|p + 1 - 4 + 1| = 1 \implies |p - 2| = 1$
This yields two values for p$p$:
- p - 2 = 1 implies p = 3$p - 2 = 1 \implies p = 3$
- p - 2 = -1 implies p = 1$p - 2 = -1 \implies p = 1$
Given that a, b$a, b$ are the parameters with a < b$a < b$, we assign a = 1$a = 1$ and b = 3$b = 3$.
### Step 2: Computing Latus Rectum of the Ellipse
The ellipse equation is:
fracx^21^2 + fracy^23^2 = 1$\frac{x^2}{1^2} + \frac{y^2}{3^2} = 1$
Since b > a$b > a$, the formula for the length of the latus rectum is:
textLatus Rectum = frac2a^2b = frac2(1)^23 = frac23$\text{Latus Rectum} = \frac{2a^2}{b} = \frac{2(1)^2}{3} = \frac{2}{3}$
### Pattern Recognition
Be careful with coordinate geometry variables; when an ellipse satisfies b > a$b > a$, the major axis is along the y-axis, making the latus rectum equal to frac2a^2b$\frac{2a^2}{b}$ instead of frac2b^2a$\frac{2b^2}{a}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 11 Mathematics: Conic Sections
Q692025Angle Between Two Lines
Let A be the point of intersection of the lines L _ 1: frac x - 71 = frac y - 50 = frac z - 3- 1$L _ {1}: \frac {x - 7}{1} = \frac {y - 5}{0} = \frac {z - 3}{- 1}$ and mathrmL_2:fracmathrmx - 13 = fracmathrmy + 34 = fracmathrmz + 75$\mathrm{L}_2:\frac{\mathrm{x} - 1}{3} = \frac{\mathrm{y} + 3}{4} = \frac{\mathrm{z} + 7}{5}$. Let B and C be the point on the lines mathrmL_1$\mathrm{L}_1$ and mathrmL_2$\mathrm{L}_2$ respectively such that mathrmAB = mathrmAC = sqrt15$\mathrm{AB} = \mathrm{AC} = \sqrt{15}$. Then the square of the area of the triangle ABC is :
A.54$54$
B.63$63$
C.57$57$
D.60$60$
Solution
### Core Logic
First, find the point of intersection A$A$ by solving the lines. Any point on L_1$L_1$ can be written as (lambda + 7, 5, -lambda + 3)$(\lambda + 7, 5, -\lambda + 3)$.
Substituting this point into the equation for L_2$L_2$:
frac(lambda + 7) - 13 = frac5 + 34 implies fraclambda + 63 = 2 implies lambda = 0$\frac{(\lambda + 7) - 1}{3} = \frac{5 + 3}{4} \implies \frac{\lambda + 6}{3} = 2 \implies \lambda = 0$
Thus, the intersection point is A = (7, 5, 3)$A = (7, 5, 3)$.
### Step 1: Calculating the Angle between lines
The directional vectors of lines L_1$L_1$ and L_2$L_2$ are vecu = hati - hatk$\vec{u} = \hat{i} - \hat{k}$ and vecv = 3hati + 4hatj + 5hatk$\vec{v} = 3hat{i} + 4hat{j} + 5hat{k}$ respectively.
costheta = frac|vecu cdot vecv||vecu||vecv| = frac|1(3) + 0(4) - 1(5)|sqrt1^2+(-1)^2 sqrt3^2+4^2+5^2 = frac|3 - 5|sqrt2sqrt50 = frac210 = frac15$\cos\theta = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}||\vec{v}|} = \frac{|1(3) + 0(4) - 1(5)|}{\sqrt{1^2+(-1)^2} \sqrt{3^2+4^2+5^2}} = \frac{|3 - 5|}{\sqrt{2}\sqrt{50}} = \frac{2}{10} = \frac{1}{5}$
Now, find sintheta$\sin\theta$:
sintheta = sqrt1 - cos^2theta = sqrt1 - frac125 = fracsqrt245$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{1}{25}} = \frac{\sqrt{24}}{5}$Three dimensional geometry diagram for Q69 - JEE Main 2025 Evening
### Step 2: Finding Area of the Triangle
The area of triangle ABC$\triangle ABC$ given two sides and their included angle is:
textArea = frac12 cdot AB cdot AC cdot sintheta$\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin\theta$
Given AB = AC = sqrt15$AB = AC = \sqrt{15}$:
textArea = frac12 cdot sqrt15 cdot sqrt15 cdot fracsqrt245 = frac15sqrt2410 = frac3sqrt242$\text{Area} = \frac{1}{2} \cdot \sqrt{15} \cdot \sqrt{15} \cdot \frac{\sqrt{24}}{5} = \frac{15\sqrt{24}}{10} = \frac{3\sqrt{24}}{2}$
Squaring the area:
textArea^2 = left(frac3sqrt242right)^2 = frac9 times 244 = 9 times 6 = 54$\text{Area}^2 = \left(\frac{3\sqrt{24}}{2}\right)^2 = \frac{9 \times 24}{4} = 9 \times 6 = 54$
### Pattern Recognition
Since B$B$ and C$C$ lie on lines intersecting at A$A$, you don't need to determine their exact coordinates to find the area of the triangle. The standard side-angle-side area formula works perfectly using just the directional angle.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 11 Mathematics: Properties of Triangles
Q562025Shortest Distance Between Two Lines
Let the shortest distance between the lines fracx - 33 = fracy - alpha-1 = fracz - 31$\frac{x - 3}{3} = \frac{y - \alpha}{-1} = \frac{z - 3}{1}$ and fracx + 3-3 = fracy + 72 = fracz - beta4$\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - \beta}{4}$ be 3sqrt30$3\sqrt{30}$. Then the positive value of 5alpha + beta$5\alpha + \beta$ is
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