Q.Find the vector and cartesian equations of the plane passing through the points having position vectors $\hat{i} + \hat{j} - 2\hat{k}$, $2\hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$. Write the equation of a plane passing through a point $(2, 3, 7)$ and parallel to the plane obtained above. Hence, find the distance between the two parallel planes.

Showing the 12 most recent of 41 on this concept.

1. CBSE 20261 markMCQ
   Q.The length of perpendicular drawn from point (2, 5, 7) on line x 1 = y 0 = z 0 is 1 (A) 2 (B) 5 (C) 74 (D) 78

   ›Reveal solutionSolution

   The perpendicular distance from a point to a line in 3D is found by projecting the vector from a point on the line to the given point onto the direction vector, then using the Pythagorean theorem. For point $(2,5,7)$ and the line $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$, the distance is $\sqrt{74}$, so the correct option is (C).

   Concept and Intuition

   The distance from a point to a line in 3D is the length of the perpendicular segment connecting the point to the line. This is not the same as the distance along any slant path — it's the shortest possible distance.

   Think of it this way: if you stand at a point in space and look at a line, the shortest path to reach that line is to walk straight towards it at a right angle. That perpendicular distance is what we calculate.

   The key idea: take any point $A$ on the line, form the vector $\overrightarrow{AP}$ from $A$ to the given point $P$, then project $\overrightarrow{AP}$ onto the direction vector $\vec{d}$ of the line. The component of $\overrightarrow{AP}$ perpendicular to $\vec{d}$ gives the perpendicular distance.

   Distance from point $P$ to line through $A$ with direction $\vec{d}$:

   $d = \frac{|\overrightarrow{AP} \times \vec{d}|}{|\vec{d}|}$

   This works because the cross product magnitude gives the area of the parallelogram formed by $\overrightarrow{AP}$ and $\vec{d}$, and dividing by $|\vec{d}|$ gives the height (perpendicular distance) of that parallelogram.

   Step-by-Step Solution

   1. Identify the line and a point on it.

   The line is given as $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$. This means:

   • Direction ratios are $(1, 0, 0)$ — the line runs along the x-axis.
   • The line passes through the origin $(0, 0, 0)$ because when $x=0$, $y=0$, $z=0$ satisfies the equation.

   So we have:

   • Point on line: $A = (0, 0, 0)$
   • Direction vector: $\vec{d} = (1, 0, 0)$
   • Given point: $P = (2, 5, 7)$

   2. Find the vector from $A$ to $P$.

   $\overrightarrow{AP} = P - A = (2-0, 5-0, 7-0) = (2, 5, 7)$

   3. Compute the cross product $\overrightarrow{AP} \times \vec{d}$.

   $\overrightarrow{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & 7 \\ 1 & 0 & 0 \end{vmatrix}$

   Expanding:

   • $\hat{i}$ component: $(5)(0) - (7)(0) = 0$
   • $\hat{j}$ component: $-( (2)(0) - (7)(1) ) = -(0 - 7) = 7$
   • $\hat{k}$ component: $(2)(0) - (5)(1) = -5$

   So $\overrightarrow{AP} \times \vec{d} = (0, 7, -5)$

   Tip

   Notice that since $\vec{d} = (1,0,0)$ is along the x-axis, the cross product simply picks out the y and z components of $\overrightarrow{AP}$ with a sign swap. This is a shortcut: for a line along the x-axis, the perpendicular distance is just $\sqrt{y^2 + z^2}$ of the point relative to the line.

   4. Find the magnitude of the cross product.

   $|\overrightarrow{AP} \times \vec{d}| = \sqrt{0^2 + 7^2 + (-5)^2} = \sqrt{0 + 49 + 25} = \sqrt{74}$ …

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2. CBSE 2026Set 65/1/11 markMCQ
   Q.The length of the perpendicular from the point $(2, 5, 7)$ on the line $\dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{0}$ is (A) $2$ (B) $5$ (C) $\sqrt{74}$ (D) $\sqrt{78}$

   ›Reveal solutionSolution

   To find the perpendicular distance from a point to a line, we identify a general point on the line, form a vector from the given point to this general point, and use the condition that this vector must be perpendicular to the line's direction vector. This allows us to find the specific point on the line (the foot of the perpendicular) and then calculate the distance. The length of the perpendicular is $\boxed{\sqrt{74}}$.

   When we talk about the "length of the perpendicular from a point to a line," we are essentially looking for the shortest distance between that point and any point on the line. Imagine dropping a plumb line from the given point straight down to the line; the length of that plumb line is what we need to find.

   The core idea is that the shortest distance occurs along a line segment that is perpendicular to the given line. If we can find the exact point on the line where this perpendicular meets it (often called the "foot of the perpendicular"), then calculating the distance between the two points becomes straightforward using the standard 3D distance formula.

   Here's how we approach this:

   1. Represent a general point on the line: Any point on the given line can be expressed using a single parameter.
   2. Form a vector: Create a vector connecting the given point to this general point on the line.
   3. Apply perpendicularity: The key insight is that this connecting vector must be perpendicular to the direction vector of the line. In 3D geometry, two vectors are perpendicular if and only if their dot product is zero. This condition will allow us to find the specific value of the parameter.
   4. Find the foot of the perpendicular: Substitute the parameter value back into the general point's coordinates to get the coordinates of the foot of the perpendicular.
   5. Calculate the distance: Use the distance formula between the given point and the foot of the perpendicular.

   Let's apply this method to the given problem.

   1. Identify the given point and the line's properties.

      The given point is $P = (2, 5, 7)$.

      The equation of the line is $\dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{0}$.

      This symmetric form tells us two crucial things:

      • A point on the line (when $x=0, y=0, z=0$) is $A = (0, 0, 0)$.
      • The direction vector of the line, $\vec{d}$, has components given by the denominators: $\vec{d} = (1, 0, 0)$.
      Note

      The line $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ is a special case. It represents the x-axis itself, as it passes through the origin $(0,0,0)$ and has a direction along the x-axis.

   2. Represent a general point on the line.

      Let $Q$ be any general point on the line. Using the parametric form of the line, $x = 0 + 1\lambda$, $y = 0 + 0\lambda$, $z = 0 + 0\lambda$, where $\lambda$ is a scalar parameter.

      So, a general point on the line is $Q = (\lambda, 0, 0)$.

   3. Form the vector connecting the given point to the general point on the line.

      The vector $\vec{PQ}$ connects point $P(2, 5, 7)$ to point $Q(\lambda, 0, 0)$.

      $\vec{PQ} = Q - P = (\lambda - 2, 0 - 5, 0 - 7) = (\lambda - 2, -5, -7)$.

   4. Apply the perpendicularity condition to find $\lambda$.

      For $\vec{PQ}$ to be the perpendicular from $P$ to the line, $\vec{PQ}$ must be perpendicular to the direction vector of the line, $\vec{d} = (1, 0, 0)$.

      The dot product of two perpendicular vectors is zero.

      If two vectors $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$ are perpendicular, then their dot product is zero:

      $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 = 0$

      So, $\vec{PQ} \cdot \vec{d} = 0$:

      $(\lambda - 2)(1) + (-5)(0) + (-7)(0) = 0$

      $\lambda - 2 + 0 + 0 = 0$

      $\lambda - 2 = 0$

      $\lambda = 2$.

   5. Find the coordinates of the foot of the perpendicular. …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.A line passing through $(2, -1, 3)$ has direction ratio (d.r.) $(3, -1, 2)$, then its equation is

   (a) $\dfrac{x+2}{3} = \dfrac{y-1}{-1} = \dfrac{z-3}{2}$

   (b) $\dfrac{x+2}{3} = \dfrac{y+1}{-1} = \dfrac{z-3}{2}$

   (c) $\dfrac{x-2}{3} = \dfrac{y+1}{-1} = \dfrac{z-3}{2}$

   (d) None of these

   ›Reveal solutionSolution

   A line through point $(x_1,y_1,z_1)$ with direction ratios $(a,b,c)$ has equation $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$.

   Here $(x_1,y_1,z_1)=(2,-1,3)$ and $(a,b,c)=(3,-1,2)$.

   …

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4. CBSE 2026Set ANNUAL1 mark
   Q.Find the vector equation of the line passing through the point (2, 3, 4) and parallel to the vector 2î + 5ĵ − 3k̂.

   ›Reveal solutionSolution

   The vector equation of a line through a point with position vector $\vec a$, parallel to $\vec b$, is $\vec r = \vec a + \lambda \vec b$.

   Position vector of the given point: $\vec a = 2\hat i+3\hat j+4\hat k$.

   Direction vector: $\vec b = 2\hat i+5\hat j-3\hat k$.

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Distance of the point $(3, -5)$ from the line $3x - 4y - 26 = 0$ is

   (a) $\dfrac{5}{3}$

   (b) $\dfrac{3}{5}$

   (c) 0

   (d) None of these

   ›Reveal solutionSolution

   Apply $d = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ with $A=3, B=-4, C=-26$ and the point $(3,-5)$.

   Distance from a point $(x_1,y_1)$ to the line $Ax+By+C=0$:

   $d = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

   Here $A=3, B=-4, C=-26$, and $(x_1,y_1)=(3,-5)$: …

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6. CBSE 2025Set 65/2/11 markMCQ
   Q.The equation of a line parallel to the vector $3\hat{i}+\hat{j}+2\hat{k}$ and passing through the point $(4,-3,7)$ is: (A) $x = 4t+3,\ y = -3t+1,\ z = 7t+2$ (B) $x = 3t+4,\ y = t+3,\ z = 2t+7$ (C) $x = 3t+4,\ y = t-3,\ z = 2t+7$ (D) $x = 3t+4,\ y = -t+3,\ z = 2t+7$

   ›Reveal solutionSolution

   The equation of a line is determined by a point it passes through and a vector parallel to it. We use the vector equation $\vec{r} = \vec{a} + t\vec{b}$ and convert it to parametric Cartesian form to find the correct option. The equation is $\boxed{x = 3t+4,\ y = t-3,\ z = 2t+7}$.

   To find the equation of a line in 3D space, we need two fundamental pieces of information:

   1. A point through which the line passes.
   2. A vector that is parallel to the line, which defines its direction.

   Imagine you are standing at a specific point in space. To define a unique line, you then need to know which way to walk. That "way to walk" is given by the direction vector. Any point on the line can be reached by starting at your initial point and moving some distance (which can be positive, negative, or zero) along the direction vector.

   Let $\vec{a}$ be the position vector of the known point $(x_1, y_1, z_1)$ through which the line passes. So, $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.

   Let $\vec{b}$ be the vector parallel to the line, which is the direction vector. So, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.

   Let $\vec{r}$ be the position vector of any arbitrary point $(x, y, z)$ on the line. So, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

   The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

   $\vec{r} = \vec{a} + t\vec{b}$

   where $t$ is a scalar parameter.

   This equation states that to reach any point $\vec{r}$ on the line, you start at $\vec{a}$ and add a scalar multiple ($t$) of the direction vector $\vec{b}$. As $t$ varies over all real numbers, $\vec{r}$ traces out all points on the line.

   Let's apply this concept to the given problem.

   1. Identify the given information.

      The line passes through the point $(4, -3, 7)$. The position vector of this point is $\vec{a} = 4\hat{i} - 3\hat{j} + 7\hat{k}$.

      The line is parallel to the vector $3\hat{i} + \hat{j} + 2\hat{k}$. This is our direction vector, $\vec{b} = 3\hat{i} + \hat{j} + 2\hat{k}$.

   2. Formulate the vector equation of the line.

      Using the formula $\vec{r} = \vec{a} + t\vec{b}$, we substitute the identified vectors:

      $\vec{r} = (4\hat{i} - 3\hat{j} + 7\hat{k}) + t(3\hat{i} + \hat{j} + 2\hat{k})$

   3. Convert the vector equation to parametric Cartesian form.

      We know that $\vec{r}$ represents any point $(x, y, z)$ on the line, so $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

      Substitute this into the equation and group the $\hat{i}$, $\hat{j}$, and $\hat{k}$ components:

      $x\hat{i} + y\hat{j} + z\hat{k} = (4\hat{i} - 3\hat{j} + 7\hat{k}) + (3t\hat{i} + t\hat{j} + 2t\hat{k})$

      $x\hat{i} + y\hat{j} + z\hat{k} = (4 + 3t)\hat{i} + (-3 + t)\hat{j} + (7 + 2t)\hat{k}$ …

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7. CBSE 2025Set 65/2/11 markMCQ
   Q.The line $x = 1 + 5\mu,\ y = -5 + \mu,\ z = -6 - 3\mu$ passes through which of the following point? (A) $(1, -5, 6)$ (B) $(1, 5, 6)$ (C) $(1, -5, -6)$ (D) $(-1, -5, 6)$

   ›Reveal solutionSolution

   To check if a point lies on a line given by parametric equations, substitute the point's coordinates into the equations and verify if a single, consistent value of the parameter $\mu$ is obtained for all three coordinates. The point $(1, -5, -6)$ yields $\mu=0$ for all equations, so it lies on the line.

   Concept and Intuition

   A line in three-dimensional space can be described using parametric equations. These equations express the $x, y,$ and $z$ coordinates of any point on the line in terms of a single parameter, often denoted by $\mu$ (or $t, \lambda$, etc.).

   The given equations are:

   $x = 1 + 5\mu$

   $y = -5 + \mu$

   $z = -6 - 3\mu$

   This means that as $\mu$ varies over all real numbers, the point $(x, y, z)$ traces out the entire line. Each specific value of $\mu$ corresponds to a unique point on the line.

   For a given point $(x_0, y_0, z_0)$ to lie on this line, there must exist one specific value of the parameter $\mu$ such that when this $\mu$ is substituted into all three equations, it simultaneously produces $x_0, y_0,$ and $z_0$. If we substitute the coordinates of a candidate point into the equations and solve for $\mu$ from each equation, we must get the same value of $\mu$ from all three equations. If the $\mu$ values are different, the point does not lie on the line.

   Step-by-Step Solution

   1. Understand the condition for a point to be on the line:

      A point $(x_0, y_0, z_0)$ lies on the line $x = 1 + 5\mu,\ y = -5 + \mu,\ z = -6 - 3\mu$ if and only if there exists a single real value of $\mu$ that satisfies all three equations simultaneously when $x=x_0, y=y_0, z=z_0$.

   2. Test Option (A): $(1, -5, 6)$

      Substitute $x=1, y=-5, z=6$ into the parametric equations:

      • For $x$: $1 = 1 + 5\mu \implies 5\mu = 0 \implies \mu = 0$
      • For $y$: $-5 = -5 + \mu \implies \mu = 0$
      • For $z$: $6 = -6 - 3\mu \implies 12 = -3\mu \implies \mu = -4$ Since the values of $\mu$ obtained are $0, 0,$ and $-4$, they are not consistent. Therefore, the point $(1, -5, 6)$ does not lie on the line.

   3. Test Option (B): $(1, 5, 6)$

      Substitute $x=1, y=5, z=6$ into the parametric equations:

      • For $x$: $1 = 1 + 5\mu \implies 5\mu = 0 \implies \mu = 0$
      • For $y$: $5 = -5 + \mu \implies \mu = 10$ The values of $\mu$ obtained are $0$ and $10$, which are not consistent. There is no need to check the $z$-coordinate. Therefore, the point $(1, 5, 6)$ does not lie on the line. …

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8. CBSE 2025Set X11 markMCQ
   Q.The equation of $y$-axis in space is

   (a) $x = 0,\ y = 0$

   (b) $x = 0,\ z = 0$

   (c) $y = 0,\ z = 0$

   (d) $y = 0$

   ›Reveal solutionSolution

   Coordinate axis as intersection of two planes — correct option (b).

   A point lies on the $y$-axis exactly when its $x$ and $z$ coordinates are both zero, with $y$ arbitrary. Henc …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.The cartesian equation of the line passing through point $(1,2,3)$ and parallel to the line $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$ will be -

   (a) $\dfrac{x-1}{3}=\dfrac{y-2}{5}=\dfrac{z-3}{6}$

   (b) $\dfrac{x+1}{3}=\dfrac{y+2}{5}=\dfrac{z+3}{6}$

   (c) $\dfrac{x+3}{1}=\dfrac{y-4}{2}=\dfrac{z+8}{3}$

   (d) $\dfrac{x+2}{3}=\dfrac{y-6}{5}=\dfrac{z+5}{6}$

   ›Reveal solutionSolution

   Parallel lines share the same direction ratios; only the point through which the line passes changes.

   The given line $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$ has direction ratios $(3,5,6)$.

   A line parallel to it and passing through $(1,2,3)$ has the same direction ratios: …

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10. CBSE 2025Set ANNUAL1 mark
    Q.Find the vector equation of the line passing through the points $(3,2,0)$ and $(1,2,5)$.

    ›Reveal solutionSolution

    The vector equation of a line through two points $A$ and $B$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$.

    Let $A(3,2,0)$ and $B(1,2,5)$, so $\vec{a}=3\hat{i}+2\hat{j}+0\hat{k}$ and $\vec{b}=1\hat{i}+2\hat{j}+5\hat{k}$.

    $\vec{b}-\vec{a}=(1-3)\hat{i}+(2-2)\hat{j}+(5-0)\hat{k}=-2\hat{i}+0\hat{j}+5\hat{k}$

    So the vector equation of the line is: …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.The vector equation of the $x$-axis is

    (a) $\vec{r} = \hat{i}$

    (b) $\vec{r} = \hat{j} + \hat{k}$

    (c) $\vec{r} = \lambda\hat{i}$

    (d) none of these

    ›Reveal solutionSolution

    The x-axis consists of all points of the form (λ, 0, 0), which as a position vector is simply λî.

    A point on the x-axis has coordinates $(\lambda, 0, 0)$ for some real λ. As a position vector this is:

    $\vec r = \lambda\hat i + 0\hat j + 0\hat k = \lambda \hat i$

    …

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12. CBSE 2025Set ANNUAL1 mark
    Q.Find the vector equation of the line through the points A(3, 4, −7) and B(1, −1, 6).

    ›Reveal solutionSolution

    The vector equation of a line through two given points $A$ and $B$ is $\vec r = \vec a + \lambda(\vec b - \vec a)$, where $\vec a, \vec b$ are the position vectors of $A, B$.

    Given: $A(3, 4, -7)$, $B(1, -1, 6)$

    Step 1 — position vectors:

    $\vec a = 3\hat i + 4\hat j - 7\hat k, \qquad \vec b = \hat i - \hat j + 6\hat k$

    Step 2 — direction vector $\vec b - \vec a$:

    $\vec b - \vec a = (1-3)\hat i + (-1-4)\hat j + (6-(-7))\hat k = -2\hat i - 5\hat j + 13\hat k$

    …

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