Q.The vector equation of the line $\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$ is __________.

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2\hat{j} - \hat{k}$ meets the line $\dfrac{x-3}{-1} = \dfrac{y+2}{5} = \dfrac{z-2}{-4}$ at $A$ and the plane $2x - y + 2z + 7 = 0$ at $B$. Then $AB =$ (A) $\sqrt{6}$ (B) $2\sqrt{6}$ (C) $3\sqrt{6}$ (D) $4\sqrt{6}$

   ›Reveal solutionSolution

   Parametrizing the line and solving for its intersections with the given line and the given plane locates $A=(2,3,-2)$, $B=(4,7,-4)$, giving $AB=2\sqrt6$.

   Concept and Intuition

   A line in space is fully described by a point and a direction vector; every point on it is found by a single parameter $t$. Intersecting it with another line means finding a common point (solve simultaneously for both parameters); intersecting it with a plane means substituting the parametrized coordinates into the plane equation and solving for $t$.

   Step-by-Step Solution

   1. Parametrize the line through $(1,1,-1)$ with direction $(1,2,-1)$: $(x,y,z)=(1+t,\,1+2t,\,-1-t)$.
   2. Parametrize the second line as $(3-s,\,-2+5s,\,2-4s)$ (from $\frac{x-3}{-1}=\frac{y+2}5=\frac{z-2}{-4}=s$).
   3. Equating: $1+t=3-s \Rightarrow t+s=2$; $1+2t=-2+5s \Rightarrow 2t-5s=-3$; $-1-t=2-4s\Rightarrow t=4s-3$.
   4. From the first, $s=2-t$; substitute into the third: $t=4(2-t)-3=5-4t \Rightarrow 5t=5 \Rightarrow t=1$, so $s=1$ (checked consistent with the second equation).
   5. So $A=(1+1,\,1+2,\,-1-1)=(2,3,-2)$. …

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2. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.The point of intersection of the lines $\bar{r} = 2\bar{b} + t(6\bar{c} - \bar{a})$ and $\bar{r} = \bar{a} + s(\bar{b} - 3\bar{c})$ is (A) $\bar{a} + \bar{b} + \bar{c}$ (B) $\bar{b} - \bar{c} - 6\bar{a}$ (C) $2\bar{a} - \bar{b} + \bar{c}$ (D) $\bar{a} + 2\bar{b} - 6\bar{c}$

   ›Reveal solutionSolution

   Matching coefficients of the (linearly independent) position vectors $\bar a,\bar b,\bar c$ on both sides of the line equations pins down the parameters and gives the intersection point $\bar a+2\bar b-6\bar c$.

   Concept and Intuition

   When two vector lines are given in terms of a common set of independent reference vectors, the intersection point can be found by comparing the coefficients of each reference vector on both sides — this is valid because $\bar a,\bar b,\bar c$ are linearly independent (as position vectors of non-collinear/non-coplanar reference points), so a vector equation between them is only satisfied if each coefficient matches independently.

   Step-by-Step Solution

   1. Line 1: $\bar r = 2\bar b + t(6\bar c-\bar a) = -t\bar a + 2\bar b + 6t\bar c$.
   2. Line 2: $\bar r = \bar a + s(\bar b-3\bar c) = \bar a + s\bar b - 3s\bar c$.
   3. Equate coefficients of $\bar a$: $-t = 1 \Rightarrow t=-1$.
   4. Equate coefficients of $\bar b$: $2 = s \Rightarrow s=2$. …

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3. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The point of intersection of the lines represented by $\bar{r}=(\bar{i}-6\bar{j}+2\bar{k})+t(\bar{i}+2\bar{j}+\bar{k})$ and $\bar{r}=(4\bar{j}+\bar{k})+s(2\bar{i}+\bar{j}+2\bar{k})$ is (A) $8\bar{i}+9\bar{j}+10\bar{k}$ (B) $8\bar{i}+8\bar{j}+7\bar{k}$ (C) $8\bar{i}+9\bar{j}+8\bar{k}$ (D) $8\bar{i}+8\bar{j}+9\bar{k}$

   ›Reveal solutionSolution

   Solving the three coordinate equations for the two line parameters (and verifying consistency) locates the intersection point as $(8,8,9)$.

   Concept and Intuition

   Two lines in space intersect only if there's a common point — that is, values of the two parameters $t$ and $s$ that make all three coordinates match simultaneously. With two unknowns and three equations, the system is over-determined; solving two of the equations for $t,s$ and then checking the third confirms genuine intersection (as opposed to skew lines).

   Step-by-Step Solution

   1. Line 1: $\bar r = (\bar i - 6\bar j+2\bar k)+t(\bar i+2\bar j+\bar k)$, giving coordinates $(1+t,\ -6+2t,\ 2+t)$.
   2. Line 2: $\bar r=(4\bar j+\bar k)+s(2\bar i+\bar j+2\bar k)$, giving coordinates $(2s,\ 4+s,\ 1+2s)$.
   3. Equating $x$: $1+t=2s$ — (i). Equating $y$: $-6+2t=4+s$ — (ii). Equating $z$: $2+t=1+2s$ — (iii).
   4. From (i): $t=2s-1$. Substitute into (ii): $-6+2(2s-1)=4+s \Rightarrow -6+4s-2=4+s \Rightarrow 4s-8=4+s \Rightarrow 3s=12 \Rightarrow s=4$.
   5. Then $t=2(4)-1=7$.
   6. Verify with (iii): $2+t=2+7=9$ and $1+2s=1+8=9$ — consistent, so the lines genuinely intersect. …

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4. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Let $\vec{a},\vec{b},\vec{c}$ be three non-coplanar vectors. Then the point of intersection of the line joining the points $\vec{a}+\vec{b}+\vec{c}$, $\vec{a}-\vec{b}+3\vec{c}$ and the line joining the points $2\vec{a}-\vec{b}+\vec{c}$, $\vec{a}-2\vec{b}+4\vec{c}$ is (A) $2\vec{a}+4\vec{c}$ (B) $3\vec{a}-3\vec{b}+5\vec{c}$ (C) $\vec{a}-2\vec{b}+4\vec{c}$ (D) $\vec{a}-\vec{b}+3\vec{c}$

   ›Reveal solutionSolution

   Parametrize both lines in terms of the (linearly independent) basis $\vec a,\vec b,\vec c$ and match coefficients to find where they meet. Answer: (C).

   Concept and Intuition

   Since $\vec a,\vec b,\vec c$ are non-coplanar, they act like an independent coordinate basis (like $\vec i,\vec j,\vec k$), so two vectors expressed in this basis are equal only if all three coefficients match separately.

   Step-by-Step Solution

   1. Line 1 through $\vec a+\vec b+\vec c$ and $\vec a-\vec b+3\vec c$: parametrize as $L_1(t)=(\vec a+\vec b+\vec c)+t[(\vec a-\vec b+3\vec c)-(\vec a+\vec b+\vec c)]=\vec a+(1-2t)\vec b+(1+2t)\vec c$.
   2. Line 2 through $2\vec a-\vec b+\vec c$ and $\vec a-2\vec b+4\vec c$: parametrize as $L_2(s)=(2\vec a-\vec b+\vec c)+s[(\vec a-2\vec b+4\vec c)-(2\vec a-\vec b+\vec c)]=(2-s)\vec a+(-1-s)\vec b+(1+3s)\vec c$.
   3. Setting $L_1(t)=L_2(s)$ and matching the $\vec a$-coefficients: $1=2-s\Rightarrow s=1$.
   4. Matching $\vec b$-coefficients: $1-2t=-1-s=-2\Rightarrow t=\tfrac32$. …

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5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\bar{i} - 2\bar{j}$ is a point on the line parallel to the vector $2\bar{i} + \bar{k}$. If $\bar{i} + 2\bar{j}$ is a point on the plane parallel to the vectors $2\bar{j} - \bar{k}$ and $\bar{i} + 2\bar{k}$, then the point of intersection of the line and the plane is (A) $-\dfrac{1}{3}(\bar{i} + 6\bar{j} + 2\bar{k})$ (B) $\dfrac{1}{3}(\bar{i} + 6\bar{j} + 2\bar{k})$ (C) $-\dfrac{1}{3}(\bar{i} - 6\bar{j} + 2\bar{k})$ (D) $\dfrac{1}{3}(\bar{i} - 6\bar{j} + 2\bar{k})$

   ›Reveal solutionSolution

   Writing the line and plane in coordinates and substituting the line's parametric form into the plane's equation gives $t=-2/3$, landing exactly on option (A).

   Concept and Intuition

   A line "parallel to a vector through a point" and a "plane parallel to two vectors through a point" are both standard 3D-geometry objects: the line is a one-parameter family, the plane's normal is the cross product of its two direction vectors. Finding their intersection is just substituting the line's parametrization into the plane's Cartesian equation and solving for the parameter.

   Step-by-Step Solution

   1. The line passes through $P_0=\bar i-2\bar j=(1,-2,0)$ and is parallel to $2\bar i+\bar k=(2,0,1)$. Parametrize: $(x,y,z)=(1+2t,\,-2,\,t)$.
   2. The plane passes through $Q_0=\bar i+2\bar j=(1,2,0)$ and is parallel to $2\bar j-\bar k=(0,2,-1)$ and $\bar i+2\bar k=(1,0,2)$.
   3. Normal to the plane: $\vec n=(0,2,-1)\times(1,0,2)$. Compute: $n_x=2(2)-(-1)(0)=4$, $n_y=(-1)(1)-0(2)=-1$, $n_z=0(0)-2(1)=-2$. So $\vec n=(4,-1,-2)$.
   4. Plane equation: $4(x-1)-1(y-2)-2(z-0)=0\Rightarrow4x-y-2z=2$.
   5. Substitute the line's coordinates: $4(1+2t)-(-2)-2t=2\Rightarrow4+8t+2-2t=2\Rightarrow6+6t=2\Rightarrow t=-\dfrac23$.
   6. Point of intersection: $x=1+2\left(-\dfrac23\right)=1-\dfrac43=-\dfrac13$, $y=-2$, $z=-\dfrac23$. …

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6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If the line joining the points $\bar i+2\bar j$ and $\bar j-2\bar k$ intersects the plane passing through the points $2\bar i-\bar j$, $2\bar j+3\bar k$ and $\bar k-2\bar i$ at $\bar r$, then $\bar r.(\bar i+\bar j+\bar k)=$ (A) 15 (B) 5 (C) 3 (D) 7

   ›Reveal solutionSolution

   Find the plane through three given points, parametrize the given line, intersect, then dot with $(1,1,1)$. Answer: $15$.

   Concept and Intuition

   A line-meets-plane problem is mechanical once both objects are in coordinate form: get the plane's normal via a cross product of two side vectors, write the line parametrically, substitute, and solve for the parameter at the intersection.

   Step-by-Step Solution

   1. Points: $A=(2,-1,0)$, $B=(0,2,3)$, $C=(-2,0,1)$ (from $2\bar i-\bar j$, $2\bar j+3\bar k$, $\bar k-2\bar i$).
   2. $\vec{AB}=B-A=(-2,3,3)$, $\vec{AC}=C-A=(-4,1,1)$.
   3. Normal $\vec n=\vec{AB}\times\vec{AC}=\big(3\cdot1-3\cdot1,\ -[(-2)(1)-(3)(-4)],\ (-2)(1)-(3)(-4)\big)=(0,-10,10)$, simplify to $(0,-1,1)$.
   4. Plane through $A$ with this normal: $-1(y-(-1))+1(z-0)=0\Rightarrow z-y=1$. (Check $B$: $3-2=1$ ✓; $C$: $1-0=1$ ✓.)
   5. Line through $P_1=(1,2,0)$ and $P_2=(0,1,-2)$: direction $\vec d=P_2-P_1=(-1,-1,-2)$, so $\bar r(t)=(1-t,\ 2-t,\ -2t)$. …

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7. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.The point of intersection of the lines joining points $\hat{i} + 2\hat{j}, 2\hat{i} - \hat{j}$ and $-\hat{i}, 2\hat{i}$ is (A) $\dfrac{5}{3}\hat{i}$ (B) $\dfrac{3\hat{i} + \hat{j}}{5}$ (C) $\dfrac{-3}{5}\hat{i}$ (D) $\dfrac{2}{5}\hat{j}$

   ›Reveal solutionSolution

   One of the two lines is simply the $x$-axis; intersecting the other line with $y=0$ gives the point $\frac53\hat i$.

   Concept and Intuition

   When two given points share the same $y$-coordinate, the line through them is just that horizontal line — a useful shortcut that avoids solving two line equations simultaneously.

   Step-by-Step Solution

   1. Convert to Cartesian points: $\hat i+2\hat j=(1,2)$; $2\hat i-\hat j=(2,-1)$; $-\hat i=(-1,0)$; $2\hat i=(2,0)$.
   2. The second pair, $(-1,0)$ and $(2,0)$, both lie on $y=0$, so that line is exactly the $x$-axis.
   3. The first line passes through $(1,2)$ and $(2,-1)$: slope $=\dfrac{-1-2}{2-1}=-3$.
   4. Equation: $y-2=-3(x-1) \Rightarrow y = 2-3x+3 = 5-3x$.
   5. Set $y=0$ (intersection with the $x$-axis): $0=5-3x \Rightarrow x=\dfrac53$. …

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8. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.Let $O(\bar{0})$, $A(\bar{i}+2\bar{j}+\bar{k})$, $B(-2\bar{i}+3\bar{k})$, $C(2\bar{i}+\bar{j})$, $D(4\bar{k})$ are position vectors of the points O, A, B, C and D. If a line passing through A and B intersects the plane passing through O, C and D at the point R, then position vector of R is (A) $-8\bar{i} - 4\bar{j} + 7\bar{k}$ (B) $2\bar{i} + \bar{j} + \bar{k}$ (C) $-7\bar{i} - 6\bar{j} - 5\bar{k}$ (D) $3\bar{i} + 2\bar{j} - 5\bar{k}$

   ›Reveal solutionSolution

   Parametrize the line $AB$, intersect it with the plane $x=2y$ (through $O,C,D$), and get $R=-8\bar i-4\bar j+7\bar k$.

   Concept and Intuition

   A line meeting a plane is found by writing the line in parametric form, substituting into the plane's Cartesian equation, and solving for the parameter. The plane through three points $O,C,D$ (one of which is the origin) has normal $\vec{OC}\times\vec{OD}$ and passes through the origin, so its equation has zero constant term.

   Step-by-Step Solution

   1. $A=(1,2,1)$, $B=(-2,0,3)$. Line $AB$: $P(t)=A+t(B-A)=(1,2,1)+t(-3,-2,2)=(1-3t,\,2-2t,\,1+2t)$.
   2. $C=(2,1,0)$, $D=(0,0,4)$. Plane through $O,C,D$ is spanned by $\vec{OC}=(2,1,0)$ and $\vec{OD}=(0,0,4)$; normal $=\vec{OC}\times\vec{OD}=\begin{vmatrix}\bar i&\bar j&\bar k\\2&1&0\\0&0&4\end{vmatrix}=(4,-8,0)$.
   3. Plane equation (through origin): $4x-8y=0\Rightarrow x=2y$. …

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9. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Points P and Q are given by $\overline{OP} = \bar{i} - \bar{j} - \bar{k}$ and $\overline{OQ} = -\bar{i} + \bar{j} + \bar{k}$. A line along the vector $\bar{a} = \bar{i} + \bar{j}$ passes through the point P and another line along the vector $\bar{b} = \bar{j} - \bar{k}$ passes through the point Q. If a line along the vector $\bar{c} = \bar{i} - \bar{j} + \bar{k}$ intersects both the lines along the vectors $\bar{a}$ and $\bar{b}$ at L and M respectively, then $\overline{PM} =$ (A) $\bar{i} - \bar{j} + 2\bar{k}$ (B) $4\bar{i} + 4\bar{j}$ (C) $-2\bar{i} + 10\bar{j} - 6\bar{k}$ (D) $3\bar{i} - 2\bar{j} + \bar{k}$

   ›Reveal solutionSolution

   Setting up the two given lines parametrically and forcing the connecting segment $LM$ to be parallel to $\bar c$ pins down the parameters, giving $\overline{PM}=-2\bar i+10\bar j-6\bar k$.

   Concept and Intuition

   When a third line (direction $\bar c$) is said to intersect two given (skew) lines, the two intersection points $L$ (on line $a$) and $M$ (on line $b$) must satisfy that the vector between them, $M-L$, is itself a scalar multiple of $\bar c$ (since both points lie on the same line of direction $\bar c$). This turns a geometric intersection condition into a simple vector equation to solve for the two free parameters.

   Step-by-Step Solution

   1. $P=(1,-1,-1)$ with $\overline{OP}=\bar i-\bar j-\bar k$. Line 1 (through $P$, direction $\bar a=\bar i+\bar j$): $L=(1+s,\,-1+s,\,-1)$ for parameter $s$.
   2. $Q=(-1,1,1)$ with $\overline{OQ}=-\bar i+\bar j+\bar k$. Line 2 (through $Q$, direction $\bar b=\bar j-\bar k$): $M=(-1,\,1+u,\,1-u)$ for parameter $u$.
   3. Since $L$ and $M$ both lie on the line of direction $\bar c=\bar i-\bar j+\bar k$, we need $M-L=k\bar c$ for some scalar $k$:

   $M-L=(-1-(1+s),\;1+u-(-1+s),\;1-u-(-1))=(-2-s,\;2+u-s,\;2-u)$

   4. Equate components to $k(1,-1,1)$: $-2-s=k$; $2+u-s=-k$; $2-u=k$.
   5. From the first and third: $-2-s=2-u\Rightarrow u-s=4$.
   6. From the second: $2+u-s=-(-2-s)=2+s\Rightarrow u-s=s\Rightarrow u=2s$. …

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10. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.If $P = (3, 12, 4)$ and Q is a point on the line OP such that $OQ = 3$ then the sum of all the coordinates of Q is (A) $\pm\dfrac{10}{13}$ (B) $\pm\dfrac{28}{13}$ (C) $\pm\dfrac{19}{13}$ (D) $\pm\dfrac{57}{13}$

    ›Reveal solutionSolution

    Scaling the unit vector along $OP$ to length $3$ (in either direction) and summing coordinates gives $\pm\dfrac{57}{13}$.

    Concept and Intuition

    Any point on line $OP$ at a fixed distance from $O$ is just a scalar multiple of the unit vector along $OP$; since $Q$ could be on either side of $O$, both signs are valid.

    Step-by-Step Solution

    1. $|OP|=\sqrt{3^2+12^2+4^2}=\sqrt{169}=13$.
    2. Unit vector along $OP$: $\hat{u}=\left(\dfrac{3}{13},\dfrac{12}{13},\dfrac{4}{13}\right)$.
    3. $Q=\pm3\hat{u}=\pm\left(\dfrac{9}{13},\dfrac{36}{13},\dfrac{12}{13}\right)$. …

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11. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.Let $\vec{a} = \hat{i}$ and $\vec{b} = \hat{j}$. The point of intersection of the lines $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is (A) $\vec{r} = \hat{i} + \hat{j}$ (B) $\vec{r} = \hat{i} - \hat{j}$ (C) $\vec{r} = \hat{k}$ (D) $\vec{r} = 2\hat{i} + \hat{j}$

    ›Reveal solutionSolution

    Each cross-product equation forces $\vec r$ onto a specific line; solving both lines simultaneously gives $\vec r=\hat i+\hat j$.

    Concept and Intuition

    An equation of the form $\vec r\times\vec c=\vec d\times\vec c$ (with $\vec d$ fixed and $\vec c$ a fixed direction) rearranges to $(\vec r-\vec d)\times\vec c=0$, meaning $\vec r-\vec d$ is parallel to $\vec c$ — i.e. $\vec r$ traces out the line through the tip of $\vec d$ in direction $\vec c$. Two such conditions together pin down a unique point: the intersection of the two lines.

    Step-by-Step Solution

    1. From $\vec r\times\vec a=\vec b\times\vec a$: $(\vec r-\vec b)\times\vec a=\vec 0$, so $\vec r-\vec b$ is parallel to $\vec a=\hat i$. Thus $\vec r=\vec b+t\hat i = \hat j + t\hat i = (t,1,0)$.
    2. From $\vec r\times\vec b=\vec a\times\vec b$: $(\vec r-\vec a)\times\vec b=\vec 0$, so $\vec r-\vec a$ is parallel to $\vec b=\hat j$. Thus $\vec r=\vec a+s\hat j=\hat i+s\hat j=(1,s,0)$.
    3. Equate the two parametrizations: $(t,1,0)=(1,s,0) \Rightarrow t=1,\ s=1$. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.Line $L_1$ passes through the points $\bar i + \bar j$ and $\bar k - \bar i$. Line $L_2$ passes through the point $\bar j + 2\bar k$ and is parallel to the vector $\bar i + \bar j + \bar k$. If $x\bar i + y\bar j + z\bar k$ is the point of intersection of the lines $L_1$ and $L_2$, then $(y - x) =$ (A) $2z$ (B) $-2z$ (C) $z$ (D) $-z$

    ›Reveal solutionSolution

    Parametrise both lines, equate coordinates to find the common point, then check the relation between $y-x$ and $z$ at that point.

    Concept and Intuition

    Two lines given by a point + direction (or two points) can be intersected by writing both as parametric equations in 3D and solving the resulting system for the two parameters — the point where they agree (if it exists) is the intersection.

    Step-by-Step Solution

    1. $L_1$ passes through $A=(1,1,0)$ ($\bar i+\bar j$) and $B=(-1,0,1)$ ($\bar k-\bar i$); direction $= B-A = (-2,-1,1)$.
    2. Parametrise: $L_1: (x,y,z) = (1-2t,\ 1-t,\ t)$.
    3. $L_2$ passes through $(0,1,2)$ ($\bar j+2\bar k$), parallel to $(1,1,1)$: $(x,y,z)=(s,\ 1+s,\ 2+s)$.
    4. Equate: $1-2t=s$, $1-t=1+s$, $t=2+s$.
    5. From the 2nd equation: $-t=s$, i.e. $s=-t$. Substitute into the 3rd: $t = 2-t \Rightarrow 2t=2 \Rightarrow t=1,\ s=-1$.
    6. Check the 1st equation: $1-2(1)=-1=s$. Consistent. …

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