Question

The equation to the altitude of the triangle formed by (1, 1, 1), (1, 2, 3), (2, -1, 1) through (1, 1, 1) is
a) 7 = (1 +1 +7)+t(i – 37 – 2K)
b) 7 = (1 + i +h)+t(31 +1 +25)
c) F= (1 + i +k) +r(7 - 1 + 2)
d) Fl=5
TE SERIES for Sri Chaitanya Jr. ICON Students
197​

Answer 1

Given :

• A =(1,1,1)

•B=(1,2,3)

•C=(2,-1,1)

and AP be the altitude

To Find :

• The equations of altitude of the triangle =?

Solution :

$\red{\bf{Given}} \begin{cases} \: \sf{A=(1,1,1)} \\ \\ \\ \sf{B=(1,2,3)} \\ \\ \\ \sf{C=(2,-1,1)} \end{cases}$

Here,

•AP be the altitude then,

$\\ \bullet\sf{DC's \: of \: AP=(\lambda ,-3 \lambda +1,-2 \lambda +2)}$

$\\ \bullet\sf{DC's \: of \: BC=(1,-3,-2)}$

$\\ \therefore\sf{P= (\lambda +1, -3 \lambda +2 , -2 \lambda +3)}$

$\\ \therefore \sf{AP \perp BC}$

$\\ \\ \implies \sf \: \lambda + 3(3 \lambda - 1) + 2(2 \lambda - 2 = 0 \\ \\ \\ \implies \sf \: \lambda + 9 \lambda - 3 + 4 \lambda - 4 = 0 \\ \\ \\ \implies \sf \: 14 \lambda - 7 = 0 \\ \\ \\ \implies \sf \: \therefore \: \lambda = \frac{7}{14} = \frac{1}{2}$

$\\ \sf{ie, \: \: \: P=\bigg(\dfrac{1}{2}+1,\dfrac{-3}{2}+2,\dfrac{-2}{2}+3 \bigg) }$

$\\ \\ \implies\sf{P= \bigg(\dfrac{3}{2},\dfrac{1}{2},2 \bigg) }$

$\\ \\ \implies\sf{AP=\dfrac{1}{2} \hat{i} -\dfrac{1}{2} \hat{j} + \hat{k}}$

$\\ \implies \sf \: r = (i + j + k) + t( i - j + 2k)$

The equations of altitude of the triangle is :

$\implies \boxed{ \sf{r =( i + j + k )+t(i - j + 2k)}}$