
In a triangle with vertices$A\left( {1, - 1,2} \right)$, $B\left( {5, - 6,2} \right)$ and$C\left( {1,3, - 1} \right)$, find the altitude $n = \left| {BD} \right|$.
A. 5
B. 10
C. $5\sqrt 2 $
D. $10\sqrt 2 $
Answer
161.7k+ views
Hint:In this given question the vertices of a triangle are already given and we need to draw an altitude from point B which will on line AC of the triangle. Also in this $\Delta ABC$ there are two perpendicular AC and BD
Complete step by step Solution:
Given
We are given a $\Delta ABC$ with the following vertices$A\left( {1, - 1,2} \right)$, $B\left( {5, - 6,2} \right)$ and$C\left( {1,3, - 1} \right)$
On this $\Delta ABC$draw an altitude on a point between AC which connects to point B and that point being point D.
Let us assume that point D has following coordinates as $\left( {l,m,n} \right)$
Now, the line ac is $\dfrac{{\left( {x - 1} \right)}}{0} = \dfrac{{\left( {y + 1} \right)}}{4} = \dfrac{{\left( {z - 2} \right)}}{{ - 3}} = t$
From the above equation we can conclude that the point D on line AC will be $\left( {1,4t - 1,2 - 3t} \right)$
Therefore, point D$\left( {l,m,n} \right)$is $\left( {1,4t - 1,2 - 3t} \right)$
According to this
$
l = 1 \\
m = 4t - 1 \\
n = 2 - 3t \\
$
BD has \[l - 5,{\text{ }}m + 6,{\text{ }}n - 2\]
AC has \[0,4, - 3\]
In this $\Delta ABC$ there are two perpendicular AC and BD, by that we can conclude that
$4(m + 6) - 3(n - 2) = 0$
$4\left( {4t + 5} \right) - 3 - \left( {3t} \right) = 0$
$t = \dfrac{{ - 4}}{5}$
From the above equation we can find out the coordinates of point D$\left( {l,m,n} \right)$
$l = 4t - 1$
\[ = 4\left( {\dfrac{{ - 4}}{5}} \right) = \dfrac{{ - 21}}{5}\]
$m = 1$
$n = 2 - 3t$
$ = 2 - 3\left( {\dfrac{{ - 4}}{5}} \right) = \dfrac{{22}}{5}$
Thus point D is $\left( {1,\dfrac{{ - 21}}{5},\dfrac{{22}}{5}} \right)$
By the above information we can say find altitude $n = \left| {BD} \right|$
$n = \left| {BD} \right| = \sqrt {{1^2} + {{\left( {\dfrac{{21}}{5}} \right)}^2} + {{\left( {\dfrac{{22}}{5}} \right)}^2}} $
$ = \sqrt {25} = 5$
Hence, the correct option is A.
Note: In this we have to draw a vertex in $\Delta ABC$ and then find out the coordinates of that point. Also, we have to assume the point coordinate and make it into an equation form which makes it easy to derive an answer. We also need to know how to find BD with the help of the coordinates given.
Complete step by step Solution:
Given
We are given a $\Delta ABC$ with the following vertices$A\left( {1, - 1,2} \right)$, $B\left( {5, - 6,2} \right)$ and$C\left( {1,3, - 1} \right)$
On this $\Delta ABC$draw an altitude on a point between AC which connects to point B and that point being point D.
Let us assume that point D has following coordinates as $\left( {l,m,n} \right)$
Now, the line ac is $\dfrac{{\left( {x - 1} \right)}}{0} = \dfrac{{\left( {y + 1} \right)}}{4} = \dfrac{{\left( {z - 2} \right)}}{{ - 3}} = t$
From the above equation we can conclude that the point D on line AC will be $\left( {1,4t - 1,2 - 3t} \right)$
Therefore, point D$\left( {l,m,n} \right)$is $\left( {1,4t - 1,2 - 3t} \right)$
According to this
$
l = 1 \\
m = 4t - 1 \\
n = 2 - 3t \\
$
BD has \[l - 5,{\text{ }}m + 6,{\text{ }}n - 2\]
AC has \[0,4, - 3\]
In this $\Delta ABC$ there are two perpendicular AC and BD, by that we can conclude that
$4(m + 6) - 3(n - 2) = 0$
$4\left( {4t + 5} \right) - 3 - \left( {3t} \right) = 0$
$t = \dfrac{{ - 4}}{5}$
From the above equation we can find out the coordinates of point D$\left( {l,m,n} \right)$
$l = 4t - 1$
\[ = 4\left( {\dfrac{{ - 4}}{5}} \right) = \dfrac{{ - 21}}{5}\]
$m = 1$
$n = 2 - 3t$
$ = 2 - 3\left( {\dfrac{{ - 4}}{5}} \right) = \dfrac{{22}}{5}$
Thus point D is $\left( {1,\dfrac{{ - 21}}{5},\dfrac{{22}}{5}} \right)$
By the above information we can say find altitude $n = \left| {BD} \right|$
$n = \left| {BD} \right| = \sqrt {{1^2} + {{\left( {\dfrac{{21}}{5}} \right)}^2} + {{\left( {\dfrac{{22}}{5}} \right)}^2}} $
$ = \sqrt {25} = 5$
Hence, the correct option is A.
Note: In this we have to draw a vertex in $\Delta ABC$ and then find out the coordinates of that point. Also, we have to assume the point coordinate and make it into an equation form which makes it easy to derive an answer. We also need to know how to find BD with the help of the coordinates given.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
