Q.Find the foot of perpendicular from the point $(2, 3, -8)$ to the line $\dfrac{4-x}{2} = \dfrac{y}{6} = \dfrac{1-z}{3}$. Also, find the perpendicular distance from the given point to the line.

Showing the 12 most recent of 46 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The perpendicular distance from the point $(1, 1, -2)$ on to the line $\dfrac{x - 1}{-1} = \dfrac{y}{2} = \dfrac{z}{1}$ is (A) $1$ (B) $2\sqrt{3}$ (C) $\sqrt{5}$ (D) $\dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   Using the cross-product formula for point-to-line distance in 3D, the perpendicular distance from $(1,1,-2)$ to the given line is $\sqrt5$.

   Concept and Intuition

   For a line through point $A$ with direction vector $\vec d$, the perpendicular distance from an external point $P$ is $\dfrac{|\vec{AP}\times\vec d|}{|\vec d|}$ — the area of the parallelogram spanned by $\vec{AP}$ and $\vec d$, divided by the base length $|\vec d|$, gives the height, which is exactly this perpendicular distance.

   Step-by-Step Solution

   1. The line $\dfrac{x-1}{-1}=\dfrac{y}{2}=\dfrac{z}{1}$ passes through $A=(1,0,0)$ with direction $\vec d=(-1,2,1)$.
   2. $P=(1,1,-2)$, so $\vec{AP}=P-A=(0,1,-2)$.
   3. $\vec{AP}\times\vec d = \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&1&-2\\-1&2&1\end{vmatrix}$ $= \mathbf i(1\cdot1-(-2)\cdot2) - \mathbf j(0\cdot1-(-2)(-1)) + \mathbf k(0\cdot2-1\cdot(-1))$ …

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2. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The perpendicular distance from the point $(-1,1,0)$ to the line joining the points $(0,2,4)$ and $(3,0,1)$ is (A) $10$ (B) $\dfrac{2\sqrt{5}}{5}$ (C) $\dfrac{5}{\sqrt{2}}$ (D) $8$

   ›Reveal solutionSolution

   This is the standard 3D point-to-line distance formula using a cross product; the perpendicular distance works out to $\dfrac{5}{\sqrt2}$.

   Concept and Intuition

   The perpendicular distance from a point $P$ to a line through $A$ with direction $\vec d$ equals $\dfrac{|\vec{AP}\times\vec d|}{|\vec d|}$: geometrically, $|\vec{AP}\times \vec d|$ is the area of the parallelogram spanned by $\vec{AP}$ and $\vec d$, and dividing by the base $|\vec d|$ gives the height, which is exactly the perpendicular distance.

   Step-by-Step Solution

   1. Take $A=(0,2,4)$ as a point on the line and direction $\vec d = (3,0,1)-(0,2,4)=(3,-2,-3)$.
   2. $\vec{AP}=P-A=(-1,1,0)-(0,2,4)=(-1,-1,-4)$.
   3. Compute $\vec{AP}\times\vec d$: $\vec{AP}\times\vec d = \big((-1)(-3)-(-4)(-2),\ -[(-1)(-3)-(-4)(3)],\ (-1)(-2)-(-1)(3)\big)$ $=(3-8,\ -(3+12),\ 2+3)=(-5,-15,5)$.
   4. $|\vec{AP}\times\vec d| = \sqrt{25+225+25}=\sqrt{275}=5\sqrt{11}$. …

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3. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Identify the point on the line $2x + 3y + 7 = 0$, which is at a distance of +3 units from $(1, -3)$. (A) $\left( \frac{\sqrt{13} + 9}{\sqrt{13}}, \frac{-3\sqrt{13} + 6}{\sqrt{13}} \right)$ (B) $\left( \frac{\sqrt{13} - 9}{\sqrt{13}}, \frac{-3\sqrt{13} - 6}{\sqrt{13}} \right)$ (C) $\left( \frac{\sqrt{13} - 9}{\sqrt{13}}, \frac{-3\sqrt{13} + 6}{\sqrt{13}} \right)$ (D) $\left( \frac{\sqrt{13} + 9}{\sqrt{13}}, \frac{3\sqrt{13} - 6}{\sqrt{13}} \right)$

   ›Reveal solutionSolution

   $(1,-3)$ already lies on the given line; moving 3 units along the line's direction locates the required point, matching option (C).

   Concept and Intuition

   A point "on the line at distance $d$ from a given point" is found by first checking that the given point lies on (or relating it to) the line, then walking along the line's direction vector by the given distance.

   Step-by-Step Solution

   1. Check $(1,-3)$ on the line $2x+3y+7=0$: $2(1)+3(-3)+7=2-9+7=0$ ✓ — it IS on the line.
   2. The line's direction vector (perpendicular to normal $(2,3)$) is $(3,-2)$, with magnitude $\sqrt{9+4}=\sqrt{13}$; unit vector $\left(\dfrac{3}{\sqrt{13}},\dfrac{-2}{\sqrt{13}}\right)$.
   3. Points at distance 3 from $(1,-3)$ along the line: $(1,-3)\pm3\left(\dfrac{3}{\sqrt{13}},\dfrac{-2}{\sqrt{13}}\right)$. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If M is the foot of the perpendicular drawn from $P(1,2,-1)$ to the plane passing through the point $A(3,-2,1)$ and perpendicular to the vector $4\bar{i} + 7\bar{j} - 4\bar{k}$, then the length of PM is (A) $\dfrac{32}{9}$ (B) $\dfrac{28}{9}$ (C) $\dfrac{26}{5}$ (D) $\dfrac{22}{5}$

   ›Reveal solutionSolution

   $PM$ is simply the perpendicular distance from $P$ to the given plane, computed via the point-to-plane distance formula: $\dfrac{28}{9}$.

   Concept and Intuition

   The foot of the perpendicular from a point to a plane is the closest point on the plane to that point, and the distance from the point to that foot equals the perpendicular distance from the point to the plane — no need to actually find $M$'s coordinates.

   Step-by-Step Solution

   1. The plane passes through $A(3,-2,1)$ and has normal vector $\bar n=(4,7,-4)$. Its equation: $4(x-3)+7(y+2)-4(z-1)=0$.
   2. Expand: $4x-12+7y+14-4z+4=0 \Rightarrow 4x+7y-4z+6=0$.
   3. Distance from $P(1,2,-1)$ to this plane: $d=\dfrac{|4(1)+7(2)-4(-1)+6|}{\sqrt{4^2+7^2+(-4)^2}}=\dfrac{|4+14+4+6|}{\sqrt{16+49+16}}=\dfrac{28}{\sqrt{81}}=\dfrac{28}{9}$. …

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5. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$P$ is a point on $x + y + 5 = 0$, whose perpendicular distance from $2x + 3y + 3 = 0$ is $\sqrt{13}$, then the coordinates of $P$ are: (A) $(20, -25)$ (B) $(1, -6)$ (C) $(-6, 1)$ (D) $(\sqrt{13}, -5-\sqrt{13})$

   ›Reveal solutionSolution

   Parametrize the point on the given line, apply the point-to-line distance formula, and solve for the parameter.

   Concept and Intuition

   Any point on $x+y+5=0$ can be written as $(t, -5-t)$. Setting its perpendicular distance to the second line equal to the given value produces a linear equation in $t$ (inside an absolute value), which we solve directly.

   Step-by-Step Solution

   1. Let $P=(t,\,-5-t)$ since $y=-5-x$ on the line $x+y+5=0$.
   2. Distance from $P$ to $2x+3y+3=0$:

   $\frac{|2t+3(-5-t)+3|}{\sqrt{2^2+3^2}} = \frac{|2t-15-3t+3|}{\sqrt{13}} = \frac{|-t-12|}{\sqrt{13}}.$

   3. Set equal to $\sqrt{13}$: $|t+12| = 13$.
   4. So $t+12=13 \Rightarrow t=1$, or $t+12=-13 \Rightarrow t=-25$. …

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6. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.The distance between a point P whose position vector is $5\bar i+\bar j+3\bar k$ and the line $\bar r = (3\bar i+7\bar j+\bar k)+t(\bar j+\bar k)$ is (A) 3 (B) 4 (C) 5 (D) 6

   ›Reveal solutionSolution

   Use the point-to-line distance formula $\dfrac{|\overline{AP}\times\bar d|}{|\bar d|}$ where $A$ is any point on the line and $\bar d$ its direction vector.

   Concept and Intuition

   The distance from a point $P$ to a line through $A$ with direction $\bar d$ equals the length of the component of $\overline{AP}$ perpendicular to the line — computed cleanly via the cross product magnitude divided by $|\bar d|$.

   Step-by-Step Solution

   1. From $\bar r = (3\bar i+7\bar j+\bar k)+t(\bar j+\bar k)$: $A=(3,7,1)$, $\bar d=(0,1,1)$.
   2. $P=(5,1,3)$, so $\overline{AP}=P-A=(2,-6,2)$.
   3. $\overline{AP}\times\bar d = \begin{vmatrix}\bar i&\bar j&\bar k\\2&-6&2\\0&1&1\end{vmatrix} = \bar i(-6\cdot1-2\cdot1)-\bar j(2\cdot1-2\cdot0)+\bar k(2\cdot1-(-6)\cdot0) = (-8,-2,2)$. …

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7. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The points $(2, 3)$ and $\left(-4, -\dfrac{4}{3}\right)$ lie on the opposite sides of the line $L \equiv 5x - 6y + k = 0$ and k is an integer. If the points $(1, 2)$ and $(4, 5)$ lie on the same side of the line $L = 0$, then the perpendicular distance from origin to the line $L = 0$ is (A) $\dfrac{7}{\sqrt{61}}$ (B) $\dfrac{9}{\sqrt{61}}$ (C) $\dfrac{10}{\sqrt{61}}$ (D) $\dfrac{11}{\sqrt{61}}$

   ›Reveal solutionSolution

   This tests the "same side / opposite side of a line" sign test (plug points into $ax+by+c$) combined with an integer constraint, then computing perpendicular distance from the origin. Answer: $\dfrac{11}{\sqrt{61}}$.

   Concept and Intuition

   For a line $L\equiv ax+by+c=0$, two points lie on opposite sides exactly when substituting their coordinates into $ax+by+c$ gives values of opposite sign (product negative); they lie on the same side when the product is positive. Using both given point-pairs narrows $k$ down to a single integer, after which the perpendicular distance from a point to a line is the standard formula $\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$.

   Step-by-Step Solution

   1. Substitute $(2,3)$: $5(2)-6(3)+k=10-18+k=k-8$.
   2. Substitute $\left(-4,-\dfrac43\right)$: $5(-4)-6\left(-\dfrac43\right)+k=-20+8+k=k-12$.
   3. Opposite sides $\Rightarrow (k-8)(k-12)<0 \Rightarrow 8<k<12$, so (with $k$ an integer) $k\in\{9,10,11\}$.
   4. Substitute $(1,2)$: $5(1)-6(2)+k=5-12+k=k-7$.
   5. Substitute $(4,5)$: $5(4)-6(5)+k=20-30+k=k-10$.
   6. Same side $\Rightarrow(k-7)(k-10)>0\Rightarrow k<7$ or $k>10$. …

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8. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.The number of straight lines that can be drawn through the point $(-3, 4)$ which are at a distance of 5 units from the point $(2, -8)$ is (A) $0$ (B) $1$ (C) $2$ (D) Infinite

   ›Reveal solutionSolution

   "Lines through a point at a fixed distance from another point" are exactly the tangent lines from that point to a circle of that radius centered at the other point. Answer: (C).

   Concept and Intuition

   A line through $(-3,4)$ at perpendicular distance exactly $5$ from $(2,-8)$ is, by definition, tangent to the circle of radius $5$ centered at $(2,-8)$. The number of tangents from a point to a circle depends on whether the point is outside (2 tangents), on (1 tangent), or inside (0 tangents) the circle.

   Step-by-Step Solution

   1. Distance between $(-3,4)$ and $(2,-8)$: $\sqrt{(2-(-3))^2+(-8-4)^2}=\sqrt{5^2+(-12)^2}=\sqrt{25+144}=\sqrt{169}=13$.
   2. We need lines through $(-3,4)$ tangent to a circle of radius $5$ centered at $(2,-8)$.
   3. Since the distance from $(-3,4)$ to the center ($13$) is greater than the radius ($5$), $(-3,4)$ lies outside the circle. …

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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The shortest distance between the lines $\bar r = \bar a + t\bar b$ and $\bar r = \bar c + s\bar d$ when $\bar a = \bar i - 2\bar j + 2\bar k$, $\bar b = 3\bar i - 2\bar j - 2\bar k$, $\bar c = 6\bar i + 2\bar j + 2\bar k$ and $\bar d = -4\bar i - \bar k$ is (A) 9 (B) $\dfrac{6\sqrt3}{\sqrt7}$ (C) $\dfrac{\sqrt7}{2\sqrt3}$ (D) $\dfrac{5}{\sqrt3}$

   ›Reveal solutionSolution

   The shortest distance between two skew lines is the projection of the vector joining a point on each line onto the common perpendicular direction $\bar b\times\bar d$; it evaluates to $\dfrac{6\sqrt3}{\sqrt7}$.

   Concept and Intuition

   Two skew lines each carry a direction vector; their cross product $\bar b\times\bar d$ points along the unique direction perpendicular to both — the direction of the shortest connecting segment. Projecting the vector between any two points on the lines onto this direction gives the shortest distance, since all other components of that connecting vector lie within the plane spanned by $\bar b,\bar d$ and contribute nothing to the perpendicular separation.

   Step-by-Step Solution

   1. $\bar a=(1,-2,2),\ \bar b=(3,-2,-2),\ \bar c=(6,2,2),\ \bar d=(-4,0,-1)$.
   2. $\bar c-\bar a=(6-1,\,2-(-2),\,2-2)=(5,4,0)$.
   3. $\bar b\times\bar d = \left((-2)(-1)-(-2)(0),\ (-2)(-4)-(3)(-1),\ (3)(0)-(-2)(-4)\right)=(2,\ 8+3,\ 0-8)=(2,11,-8)$.
   4. $|\bar b\times\bar d|=\sqrt{2^2+11^2+8^2}=\sqrt{4+121+64}=\sqrt{189}=3\sqrt{21}$.
   5. $(\bar c-\bar a)\cdot(\bar b\times\bar d)=5(2)+4(11)+0(-8)=10+44=54$. …

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10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.If a straight line L perpendicular to the line $3x-4y=6$ forms a triangle of area 6 square units with coordinate axes, then the minimum perpendicular distance from the point (1,1) to the line L is (A) $1$ (B) $\sqrt{2}$ (C) $2$ (D) $\sqrt{3}$

    ›Reveal solutionSolution

    Two lines satisfy the perpendicularity + area conditions ($4x+3y=\pm12$); the smaller of the two perpendicular distances from $(1,1)$ is $1$.

    Concept and Intuition

    A line perpendicular to a given line has the negative-reciprocal slope, which fixes its direction (here giving the family $4x+3y=k$). The area-with-axes condition pins down $|k|$, but both signs of $k$ are geometrically valid, giving two candidate lines — the question then asks for the smaller of the two resulting distances.

    Step-by-Step Solution

    1. $3x-4y=6\Rightarrow y=\dfrac{3x-6}{4}$, slope $=\dfrac34$. Perpendicular slope $=-\dfrac43$, so $L: 4x+3y=k$ for some $k$.
    2. $x$-intercept $=k/4$, $y$-intercept $=k/3$. Triangle area with axes $=\dfrac12\left|\dfrac k4\right|\left|\dfrac k3\right|=\dfrac{k^2}{24}$.
    3. Set $=6$: $k^2=144\Rightarrow k=\pm12$. …

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11. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.For a positive real number p, if the perpendicular distance from a point $-\bar{i}+p\bar{j}-3\bar{k}$ to the plane $\bar{r}\cdot(2\bar{i}-3\bar{j}+6\bar{k})=7$ is 6 units, then $p =$ (A) $\dfrac{4}{5}$ (B) $\dfrac{5}{6}$ (C) $6$ (D) $5$

    ›Reveal solutionSolution

    Applying the point-to-plane distance formula and solving for the positive root gives $p=5$.

    Concept and Intuition

    The perpendicular distance from a point $(x_0,y_0,z_0)$ to the plane $ax+by+cz=d$ is $\dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$ — a direct plug-in once the plane is written in Cartesian form.

    Step-by-Step Solution

    1. The plane $\bar r\cdot(2\bar i-3\bar j+6\bar k)=7$ is $2x-3y+6z=7$, with normal magnitude $\sqrt{4+9+36}=\sqrt{49}=7$.
    2. Point: $(-1,\ p,\ -3)$.
    3. Distance $=\dfrac{|2(-1)-3(p)+6(-3)-7|}{7} = \dfrac{|-2-3p-18-7|}{7}=\dfrac{|-27-3p|}{7}$.
    4. Set equal to 6: $|-27-3p|=42$. Since $p>0$, $-27-3p<0$, so $27+3p=42 \Rightarrow 3p=15 \Rightarrow p=5$. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.The angle made by a line L with positive X-axis measured in the positive direction is $\dfrac{\pi}{6}$ and the intercept made by L on Y-axis is negative. If L is at a distance of 5 units from the origin, then the perpendicular distance from the point $\left(1, -\sqrt{3}\right)$ to the line L is (A) $2$ (B) $1$ (C) $4$ (D) $3$

    ›Reveal solutionSolution

    Build the line from its inclination, sign of intercept, and distance from origin, then apply the point-to-line distance formula: the answer is $3$.

    Concept and Intuition

    A line's inclination fixes its slope; combined with the distance from the origin (and the sign of the intercept, which resolves the ambiguity of which parallel line it is), the line's equation is fully determined. From there, perpendicular distance from any point is a direct formula.

    Step-by-Step Solution

    1. Inclination $\pi/6 \Rightarrow$ slope $m=\tan(\pi/6)=\dfrac{1}{\sqrt3}$. Line: $y=\dfrac{x}{\sqrt3}+c$, i.e. $x-\sqrt3y+\sqrt3c=0$.
    2. Distance from origin $=\dfrac{|\sqrt3c|}{\sqrt{1+3}}=\dfrac{\sqrt3|c|}{2}=5\Rightarrow |c|=\dfrac{10}{\sqrt3}$.
    3. $y$-intercept is $c$, which must be negative, so $c=-\dfrac{10}{\sqrt3}$.
    4. Line: $x-\sqrt3y-10=0$ (multiplying through by $\sqrt3$ and simplifying, the constant term becomes $-10$). Check: distance from origin $=\dfrac{10}{\sqrt{1+3}}=5$. ✓. …

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