Q.(a) Verify that the lines given by $\vec{r} = (1-\lambda)\hat{i} + (\lambda - 2)\hat{j} + (3 - 2\lambda)\hat{k}$ and $\vec{r} = (\mu + 1)\hat{i} + (2\mu - 1)\hat{j} - (2\mu + 1)\hat{k}$ are skew lines. Hence, find the shortest distance between the lines.

Showing the 12 most recent of 13 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.The position vector of the midpoint of the line joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$

   (a) $3\hat{i}+2\hat{j}+\hat{k}$

   (b) $3\hat{i}+2\hat{j}-\hat{k}$

   (c) $\hat{i}-\hat{j}-3\hat{k}$

   (d) $-\hat{i}+\hat{j}+3\hat{k}$

   ›Reveal solutionSolution

   Averaging the coordinates of $P$ and $Q$ gives $(3,2,1)$; answer (a).

   The position vector of the midpoint is the average of the two position vectors: …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $2\vec{a}+3\vec{b}-5\vec{c}=\vec{0}$, then write the ratio in which $\vec{c}$ divides $\overline{AB}$, where the position vectors of $A$ and $B$ are respectively $\vec{a}$ and $\vec{b}$.

   (a) 3 : 2 internally

   (b) 3 : 2 externally

   (c) 2 : 3 internally

   (d) 2 : 3 externally

   ›Reveal solutionSolution

   Rearranging $2\vec a+3\vec b-5\vec c=\vec 0$ shows $\vec c$ is the point dividing $\overline{AB}$ internally in the ratio $3:2$.

   We are given $2\vec a+3\vec b-5\vec c=\vec 0$, i.e.

   $5\vec c=2\vec a+3\vec b \quad\Rightarrow\quad \vec c=\frac{2\vec a+3\vec b}{5}=\frac{3\vec b+2\vec a}{3+2}$

   Section formula: if a point $C$ divides $\overline{AB}$ internally in the ratio $m:n$ (i.e. $AC:CB=m:n$), its position vector is

   $\vec c=\frac{n\vec a+m\vec b}{m+n}$

   …

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3. CBSE 2025Set 65/4/11 markMCQ
   Q.If P is a point on the line segment joining $(3, 6, -1)$ and $(6, 2, -2)$ and y-coordinate of P is 4, then its z-coordinate is : (A) $-\frac{3}{2}$ (B) $0$ (C) $1$ (D) $\frac{3}{2}$

   ›Reveal solutionSolution

   Using the section formula in 3D, the point dividing the segment in a fixed ratio has coordinates that are weighted averages. Given the y-coordinate is 4, we find the ratio $m:n = 1:1$ (so $P$ is the midpoint) and then compute the z-coordinate as $-\frac{3}{2}$, which matches option (A).

   We have two points: $A(3, 6, -1)$ and $B(6, 2, -2)$. A point $P$ lies on the line segment $AB$, and its y-coordinate is given as $4$. We need its z-coordinate.

   The key idea is the section formula for internal division in 3D. If a point $P$ divides the segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $m:n$ (measured from $A$ to $B$), then:

   $P = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)$

   This is simply a weighted average: the coordinates of $P$ are closer to $B$ if $m > n$, and closer to $A$ if $n > m$.

   $P = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)$

   Now, we know the y-coordinate of $P$ is $4$. So:

   $\frac{m \cdot 2 + n \cdot 6}{m+n} = 4$

   Simplify:

   $2m + 6n = 4(m+n)$

   $2m + 6n = 4m + 4n$

   $6n - 4n = 4m - 2m$

   $2n = 2m$

   $m = n$

   So the ratio $m:n = 1:1$. That means $P$ is the midpoint of $AB$. …

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4. CBSE 2025Set 65/4/11 markMCQ
   Q.If the sides AB and AC of $\triangle ABC$ are represented by vectors $\hat{j} + \hat{k}$ and $3\hat{i} - \hat{j} + 4\hat{k}$ respectively, then the length of the median through A on BC is : (A) $2\sqrt{2}$ units (B) $\sqrt{18}$ units (C) $\frac{\sqrt{34}}{2}$ units (D) $\frac{\sqrt{48}}{2}$ units

   ›Reveal solutionSolution

   The median from vertex A goes to the midpoint of BC. Using the position vectors of B and C (found from the given side vectors), the median vector is half the sum of the position vectors of B and C minus the position vector of A. Its magnitude gives the length, which simplifies to $\frac{\sqrt{34}}{2}$ units.

   We are given the side vectors of $\triangle ABC$ from vertex A:

   $\overrightarrow{AB} = \hat{j} + \hat{k}$ and $\overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$.

   We need the length of the median from A to side BC.

   Concept first: A median from a vertex goes to the midpoint of the opposite side. If we place A at the origin (or treat position vectors relative to A), then the position vectors of B and C are simply $\overrightarrow{AB}$ and $\overrightarrow{AC}$. The midpoint M of BC has position vector $\frac{\overrightarrow{OB} + \overrightarrow{OC}}{2}$. The median vector is $\overrightarrow{AM} = \overrightarrow{OM} - \overrightarrow{OA}$. Since we can set A as origin, $\overrightarrow{OA} = \mathbf{0}$, so the median vector is just $\frac{\overrightarrow{OB} + \overrightarrow{OC}}{2}$. Then its length is half the magnitude of the sum of the two side vectors.

   Let’s work it through.

   1. Set A as origin.

      Let $\vec{A} = \mathbf{0}$. Then

      $\vec{B} = \overrightarrow{AB} = \hat{j} + \hat{k}$

      $\vec{C} = \overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$

   2. Find the midpoint M of BC.

      The position vector of M is

      $\vec{M} = \frac{\vec{B} + \vec{C}}{2} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2}$

      Simplify the numerator:

      $\hat{j} - \hat{j} = 0$, so the $\hat{j}$ terms cancel.

      $\hat{k} + 4\hat{k} = 5\hat{k}$

      So numerator = $3\hat{i} + 5\hat{k}$

      Hence $\vec{M} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}$

   3. The median vector from A to M.

      Since A is at origin, $\overrightarrow{AM} = \vec{M} - \vec{A} = \vec{M} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}$

   4. Length of the median. …

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5. CBSE 2024Set ANNUAL1 mark
   Q.Find the position vector of the mid-point of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

   ›Reveal solutionSolution

   The midpoint's coordinates are the average of the corresponding coordinates of the two endpoints.

   $P(2,3,4)$, $Q(4,1,-2)$.

   Midpoint $M = \left(\dfrac{2+4}{2}, \dfrac{3+1}{2}, \dfrac{4+(-2)}{2}\right) = (3, 2, 1)$

   …

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6. CBSE 2023Set 65/2/11 markMCQ
   Q.Position vector of the mid-point of line segment $AB$ is $3\hat{i} + 2\hat{j} - 3\hat{k}$. If the position vector of the point $A$ is $2\hat{i} + 3\hat{j} - 4\hat{k}$, then the position vector of the point $B$ is:

   (a) $\frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} - \frac{7}{2}\hat{k}$

   (b) $4\hat{i} + \hat{j} - 2\hat{k}$

   (c) $5\hat{i} + 5\hat{j} - 7\hat{k}$

   (d) $\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$

   ›Reveal solutionSolution

   The midpoint formula relates the position vectors of endpoints and their midpoint: $\vec{M} = \frac{\vec{A} + \vec{B}}{2}$. Rearranging gives $\vec{B} = 2\vec{M} - \vec{A}$, which yields $\vec{B} = 4\hat{i} + \hat{j} - 2\hat{k}$.

   The midpoint of a line segment is the average of its endpoints. In vector form, if $M$ is the midpoint of segment $AB$, then the position vector of $M$ is simply the arithmetic mean of the position vectors of $A$ and $B$. This comes from the fact that to reach $M$ from the origin, you can go to $A$, then travel halfway along the displacement from $A$ to $B$.

   We're given:

   • Position vector of midpoint $M$: $\vec{r}_M = 3\hat{i} + 2\hat{j} - 3\hat{k}$
   • Position vector of point $A$: $\vec{r}_A = 2\hat{i} + 3\hat{j} - 4\hat{k}$
   • Need to find: Position vector of point $B$, $\vec{r}_B$

   $\vec{r}_M = \frac{\vec{r}_A + \vec{r}_B}{2}$

   Now we solve for $\vec{r}_B$:

   1. Multiply both sides by 2 to eliminate the fraction:

   $2\vec{r}_M = \vec{r}_A + \vec{r}_B$

   2. Isolate $\vec{r}_B$ by subtracting $\vec{r}_A$ from both sides:

   $\vec{r}_B = 2\vec{r}_M - \vec{r}_A$

   3. Substitute the given vectors:

   $\vec{r}_B = 2(3\hat{i} + 2\hat{j} - 3\hat{k}) - (2\hat{i} + 3\hat{j} - 4\hat{k})$

   4. Distribute the scalar multiplication: …

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7. CBSE 2023Set 65/3/11 markMCQ
   Q.In $\triangle ABC$, $\vec{AB} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$. If $D$ is mid-point of $BC$, then vector $\vec{AD}$ is equal to :

   (a) $4\hat{i} + 6\hat{k}$

   (b) $2\hat{i} - 2\hat{j} + 2\hat{k}$

   (c) $\hat{i} - \hat{j} + \hat{k}$

   (d) $2\hat{i} + 3\hat{k}$

   ›Reveal solutionSolution

   The midpoint of a side divides the sum of the two position vectors from a vertex; here $\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$, giving $\vec{AD} = 2\hat{i} + 3\hat{k}$.

   The key insight is to express the position vector of the midpoint $D$ in terms of the vectors we already know from vertex $A$.

   When $D$ is the midpoint of $BC$, we can think of reaching $D$ from $A$ by averaging the two paths: one through $B$ and one through $C$. This is the midpoint theorem in vector form.

   To see why, imagine walking from $A$ to $B$ (vector $\vec{AB}$), then from $B$ to $D$ (vector $\vec{BD}$). Alternatively, walk from $A$ to $C$ (vector $\vec{AC}$), then from $C$ to $D$ (vector $\vec{CD}$). Since $D$ is the midpoint, $\vec{BD} = -\vec{CD}$ and both equal half of $\vec{BC}$.

   The elegant shortcut: the position vector of the midpoint from any origin is the average of the position vectors of the endpoints from that origin.

   $\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$

   Now we compute step by step:

   1. Write out the given vectors:

   $\vec{AB} = \hat{i} + \hat{j} + 2\hat{k}$

   $\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$

   2. Add the two vectors component-wise: …

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8. CBSE 2023Set ANNUAL1 mark
   Q.Find the position vector of a point R which internally divides the line joining two points P and Q whose position vectors are $(\hat{i}+2\hat{j}-\hat{k})$ and $(-\hat{i}+\hat{j}+\hat{k})$ respectively in the ratio $2:1$.

   ›Reveal solutionSolution

   Use the section formula for internal division: $\vec{r}=\dfrac{m\vec q+n\vec p}{m+n}$ for a point dividing $PQ$ in ratio $m:n$.

   $\vec p=\hat i+2\hat j-\hat k$, $\vec q=-\hat i+\hat j+\hat k$, ratio $2:1$ (R divides PQ so that $PR:RQ=2:1$).

   $\vec r=\dfrac{2\vec q+1\cdot\vec p}{2+1}=\dfrac{2(-\hat i+\hat j+\hat k)+(\hat i+2\hat j-\hat k)}{3}$

   …

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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.The $x$-axis divide the line segment joining the points $(2, -3)$ and $(5, 6)$ is

   (a) $1 : 2$

   (b) $2 : 1$

   (c) $1 : 3$

   (d) none

   ›Reveal solutionSolution

   The $x$-axis divides the segment in ratio $1:2$; option (a).

   Let the $x$-axis ($y=0$) divide the join of $(2,-3)$ and $(5,6)$ in ratio $k:1$. Using the $y$-coordinate:

   $\dfrac{6k+(-3)}{k+1}=0\Rightarrow 6k=3\Rightarrow k=\dfrac12.$

   …

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10. CBSE 2022Set ANNUAL1 markMCQ
    Q.The position vectors of the points $A$ and $B$ are $3\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}-3\hat{j}-\hat{k}$ respectively. Write the position vector of the point which divides $\overline{AB}$ in the ratio $1:3$ internally.

    (a) $\dfrac52\hat i - \dfrac74\hat k$

    (b) $\dfrac32\hat i - 2\hat j - \dfrac54\hat k$

    (c) $4\hat i + 3\hat j - \dfrac52\hat k$

    (d) $5\hat j - \dfrac12\hat k$

    ›Reveal solutionSolution

    Use the section formula for internal division: $P=\dfrac{m\vec b+n\vec a}{m+n}$ for ratio $m:n$ from $A$ to $B$.

    $\vec a=3\hat i+\hat j-2\hat k$ (point $A$), $\vec b=\hat i-3\hat j-\hat k$ (point $B$). Ratio $1:3$ (from $A$) means $m=1,\,n=3$.

    …

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11. CBSE 2020Set 65/1/11 mark
    Q.The position vectors of two points A and B are respectively $\vec{OA} = 2\hat{i} - \hat{j} - \hat{k}$ and $\vec{OB} = 2\hat{i} - \hat{j} + 2\hat{k}$. If point P divides the line segment AB in the ratio $2:1$, then its position vector is ________. Questions number 16 to 20 are Very Short Answer Type Questions.

    ›Reveal solutionSolution

    Using the section formula for internal division, the position vector of point P dividing AB in the ratio 2:1 is $\boxed{2\hat{i} - \hat{j} + \hat{k}}$.

    The section formula is the natural tool here. When a point divides a line segment in a given ratio, its position vector is a weighted average of the endpoints. For internal division, the weights are the parts of the ratio — the point is closer to the endpoint with the larger part.

    Here, P divides AB in the ratio 2:1. That means AP:PB = 2:1. Since the ratio is from A to B, P is closer to B (the larger part is from A to P, so P is 2/3 of the way from A to B). The formula gives:

    If point P divides AB internally in the ratio $m:n$ (i.e., AP:PB = m:n), then

    $\vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m+n}$

    Notice the swap: the coefficient of $\vec{OA}$ is $n$ (the opposite part) and of $\vec{OB}$ is $m$. This is because the weighted average pulls P toward the endpoint with the larger weight.

    Let’s apply it step by step.

    1. Identify the given vectors and ratio.

       $\vec{OA} = 2\hat{i} - \hat{j} - \hat{k}$

       $\vec{OB} = 2\hat{i} - \hat{j} + 2\hat{k}$

       Ratio $m:n = 2:1$, where $m$ corresponds to AP and $n$ to PB.

    2. Plug into the section formula.

    $\vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m+n} = \frac{1 \cdot (2\hat{i} - \hat{j} - \hat{k}) + 2 \cdot (2\hat{i} - \hat{j} + 2\hat{k})}{2+1}$

    3. Simplify the numerator.

       First term: $2\hat{i} - \hat{j} - \hat{k}$

       Second term: $4\hat{i} - 2\hat{j} + 4\hat{k}$

       Adding: $(2+4)\hat{i} + (-1-2)\hat{j} + (-1+4)\hat{k} = 6\hat{i} - 3\hat{j} + 3\hat{k}$

    4. Divide by the sum of the ratio parts (3). …

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12. CBSE 2020Set ANNUAL1 markMCQ
    Q.The position vector of the point which divides the join of the points $2\vec{a}-3\vec{b}$ and $\vec{a}+\vec{b}$ in the ratio of 3:1 internally is

    (a) $\dfrac{3\vec{a}-2\vec{b}}{2}$

    (b) $\dfrac{7\vec{a}-8\vec{b}}{4}$

    (c) $\dfrac{3\vec{a}}{4}$

    (d) $\dfrac{5\vec{a}}{4}$

    ›Reveal solutionSolution

    Apply the section formula for internal division: (m·P2+n·P1)/(m+n) for ratio m:n.

    Points: $P_1=2\vec a-3\vec b$, $P_2=\vec a+\vec b$; ratio $3:1$ internally means we weight $P_2$ by 3 and $P_1$ by 1: …

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