Question

The orthocentre of the triangle formed by the points (2, 1, 5), (3, 2, 3), (4, 0, 4) is______

A.(2, 1, 5)

B.(3, 2, 3)

C.(4, 0, 4)

D.(3, 0, 5)

Answer 1

Hint: Orthocentre is the point of intersection of altitudes of triangles. If we see in the below figure H is the orthocentre. We know that the equation of a line with  \[A({x_1},{y_1},{z_1})\] and  \[B({x_2},{y_2},{z_2})\]  is  \[\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}\] . We find the equation of each line in a triangle. We also know if two lies are perpendicular then the product of their direction cosines are zero.

Complete step-by-step answer:

Let,  \[A(2,1,5)\]  \[B(3,2,3)\] and  \[C(4,0,4)\] .

Now, expressing a given problem in a diagram.

Equation of a line with  \[A({x_1},{y_1},{z_1})\] and  \[B({x_2},{y_2},{z_2})\]  is  \[\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}\] 

Now, equation of line AC is given by:

 \[A(2,1,5)\] and  \[C(4,0,4)\] substituting we get,

 \[ \Rightarrow \dfrac{{x - 2}}{{4 - 2}} = \dfrac{{y - 1}}{{0 - 1}} = \dfrac{{z - 5}}{{4 - 5}}\] 

 \[ \Rightarrow \dfrac{{x - 2}}{2} = \dfrac{{y - 1}}{{ - 1}} = \dfrac{{z - 5}}{{ - 1}}\] 

Hence the direction ratios of AC is  \[(2, - 1, - 1)\] .

Equation of line BC is given by:

 \[B(3,2,3)\] and  \[C(4,0,4)\] substituting we get:

 \[ \Rightarrow \dfrac{{x - 3}}{{4 - 3}} = \dfrac{{y - 2}}{{0 - 2}} = \dfrac{{z - 3}}{{4 - 3}}\] 

 \[ \Rightarrow \dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{ - 2}} = \dfrac{{z - 3}}{1}\] 

Hence, the direction ratios of BC is  \[(1, - 2,1)\] .

We need the equation of AP.

To find P that is the foot of perpendicular BC.

We know the direction ratios of BC is  \[(1, - 2,1)\] .

 \[ \Rightarrow \dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{ - 2}} = \dfrac{{z - 3}}{1} = \lambda \] 

 \[ \Rightarrow x - 3 = \lambda ,{\text{  }}y - 2 =  - 2\lambda ,{\text{  }}z - 3 = \lambda \] 

 \[ \Rightarrow x = \lambda  + 3,{\text{  }}y =  - 2\lambda  + 2,{\text{  }}z = \lambda  + 3\] 

Then the zonal coordinates  \[(\lambda  + 3, - 2\lambda  + 2,\lambda  + 3)\] 

Now this P is perpendicular to the point A. then the direction ratios are proportional to,

 \[(\lambda  + 3 - 2, - 2\lambda  + 2 - 1,\lambda  + 3 - 5)\] 

 \[ \Rightarrow (\lambda  + 1, - 2\lambda  + 1,\lambda  - 2)\] .  - (1)

Now direction ratios of A and P are perpendicular so the sum of the product becomes zero.

  \[ \Rightarrow 1(\lambda  + 1) + ( - 2)( - 2\lambda  + 1) + 1(\lambda  - 2) = 0\] 

 \[ \Rightarrow \lambda  + 1 - 4\lambda  - 2 + \lambda  - 2 = 0\] 

 \[ \Rightarrow 6\lambda  - 3 = 0\] 

 \[ \Rightarrow \lambda  = \dfrac{3}{6}\] 

 \[ \Rightarrow \lambda  = \dfrac{1}{2}\] 

Substituting in equation (1)

 \[ \Rightarrow P = \left( {\dfrac{1}{2} + 3,{\text{ }} - 2\left( {\dfrac{1}{2}} \right) + 2,{\text{ }}\dfrac{1}{2} + 3} \right)\] 

 \[ \Rightarrow P = \left( {\dfrac{7}{2},1,\dfrac{7}{2}} \right)\] 

Now the equation of line AP with  \[A(2,1,5)\] and  \[P = \left( {\dfrac{7}{2},1,\dfrac{7}{2}} \right)\]  is :

 \[ \Rightarrow \dfrac{{x - 2}}{{2 - \dfrac{7}{2}}} = \dfrac{{y - 1}}{{1 - 1}} = \dfrac{{z - 5}}{{5 - \dfrac{7}{2}}}\] 

 \[ \Rightarrow \dfrac{{x - 2}}{{\dfrac{{ - 3}}{2}}} = \dfrac{{y - 1}}{0} = \dfrac{{z - 5}}{{\dfrac{3}{2}}}\]           (Multiply the numerator by 2)

 \[ \Rightarrow \dfrac{{x - 2}}{{ - 3}} = \dfrac{{y - 1}}{0} = \dfrac{{z - 5}}{3}\]   (divide numerator by 3)

 \[ \Rightarrow \dfrac{{x - 2}}{{ - 1}} = \dfrac{{y - 1}}{0} = \dfrac{{z - 5}}{1}\] 

Now similarly we need the equation of line BQ.

To find Q, the foot of line perpendicular to AC. 

Follow the same procedure as we did in above,

 \[ \Rightarrow \dfrac{{x - 2}}{2} = \dfrac{{y - 1}}{{ - 1}} = \dfrac{{z - 5}}{1} = \lambda \] 

 \[ \Rightarrow x = 2\lambda  + 2,{\text{  }}y =  - \lambda  + 1,{\text{  }}z =  - \lambda  + 5\] 

 \[ \Rightarrow (2\lambda  + 2, - \lambda  + 1, - \lambda  + 5)\]  - (2)

Now direction ratios are proportional to 

 \[ \Rightarrow (2\lambda  + 2 - 3, - \lambda  + 1 - 2, - \lambda  + 5 - 3)\] 

 \[ \Rightarrow (2\lambda  - 1, - \lambda  - 1, - \lambda  + 2)\] 

 \[ \Rightarrow 2(2\lambda  - 1) + ( - 1)( - \lambda  - 1) + ( - 1)( - \lambda  + 2) = 0\] 

 \[ \Rightarrow 4\lambda  - 2 + \lambda  + 1 + \lambda  - 2 = 0\] 

 \[ \Rightarrow 6\lambda  - 3 = 0\] 

 \[ \Rightarrow 6\lambda  = 3\] 

 \[ \Rightarrow \lambda  = \dfrac{3}{6}\] 

 \[ \Rightarrow \lambda  = \dfrac{1}{2}\] 

Now substituting in equation (2)

 \[\left( {2\left( {\dfrac{1}{2}} \right) + 2,{\text{ }} - \dfrac{1}{2} + 1,{\text{ }} - \dfrac{1}{2} + 5} \right)\] 

 \[Q = \left( {3,\dfrac{1}{2},\dfrac{9}{2}} \right)\] 

Now equation of BQ is 

 \[ \Rightarrow \dfrac{{x - 3}}{{3 - 3}} = \dfrac{{y - 2}}{{2 - \dfrac{1}{2}}} = \dfrac{{z - 3}}{{3 - \dfrac{9}{2}}}\] 

 \[ \Rightarrow \dfrac{{x - 3}}{0} = \dfrac{{y - 2}}{{\dfrac{3}{2}}} = \dfrac{{z - 3}}{{\dfrac{{ - 3}}{2}}}\] 

 \[ \Rightarrow \dfrac{{x - 3}}{0} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{{ - 3}}\] 

Thus we need to find the point of intersection of two lines AP and BQ:

 \[\dfrac{{x - 2}}{{ - 1}} = \dfrac{{y - 1}}{0} = \dfrac{{z - 5}}{{ - 1}}\] and  \[\dfrac{{x - 3}}{0} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{{ - 3}}\] 

Let  \[\dfrac{{x - 3}}{0} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{{ - 3}} = \lambda \] 

That is  \[H = (3,3\lambda  + 2, - 3\lambda  + 3)\] 

To find  \[\lambda \] value, 

 \[\dfrac{{x - 2}}{{ - 1}} = \dfrac{{y - 1}}{0} = \dfrac{{z - 5}}{{ - 1}} = \lambda \] 

 \[x =  - \lambda  + 2,y = 1,z =  - \lambda  + 5\] 

 \[\dfrac{{ - \lambda  + 2 - 3}}{0} = \dfrac{{1 - 2}}{3} = \dfrac{{ - \lambda  + 5 - 3}}{{ - 3}}\] 

 \[\dfrac{{ - \lambda  - 1}}{0} = \dfrac{{ - 1}}{3} = \dfrac{{ - \lambda  + 2}}{{ - 3}}\] 

 \[\dfrac{{ - \lambda  - 1}}{0} = \dfrac{{ - 1}}{3}\] 

 \[3( - \lambda  - 1) =  - 1\] 

 \[3(\lambda  + 1) = 1\] 

 \[3\lambda  + 3 = 1\] 

 \[3\lambda  =  - 2\] 

 \[\lambda  = \dfrac{{ - 2}}{3}\] 

Substituting in H.

 \[H = (3,3 \times \dfrac{{ - 2}}{3} + 2, - 3 \times \dfrac{{ - 2}}{3} + 3)\] 

 \[H = (3,0,5)\] 

So, the correct answer is “Option D”.

Note: This problem has many calculations so be careful while substituting the values. Know the equation of line passing through the two points. Also know how to find the point of intersection of two lines in three dimensions. Find the equation of line which is required as we did above.