Q.Prove that the altitudes in a triangle are concurrent.

THIS PROBLEM IS IN VECTOR ALGEBRA, MATH CHAPTER

Here in triangle ABC , we have CM perpendicular to AB and AK perpendicular to BC
And AK and CM are intersecting at H .
And BH will intersect at L ,if BL is perpendicular to AC.

So we have to prove that BL is perpendicular to AC .

$$\begin{gathered} \mathrm{Let} H \mathrm{be} \mathrm{the} \mathrm{origin} \mathrm{and} \mathrm{let} \mathrm{the} \mathrm{position} \mathrm{vectors} \mathrm{of} A = \overrightarrow{a} , B =\overrightarrow{b} , C =\overrightarrow{c} \\ \mathrm{Then} HA = \overrightarrow{a} , HB =\overrightarrow{b} , HC\hspace{0.17em}= \overrightarrow{c} \\ \mathrm{Now} AK \mathrm{is} \mathrm{perpendicular} \mathrm{to} BC, \mathrm{we} \mathrm{have} HA \mathrm{perpendicular} \mathrm{to} BC \\ \mathrm{Hence} \overrightarrow{a}.(\overrightarrow{c}-\overrightarrow{b}) =0 \left(1\right) \\ \mathrm{Similarly} HC \mathrm{is} \mathrm{perpendicular} \mathrm{to} AB \\ \mathrm{Hence} \overrightarrow{c}.(\overrightarrow{b}-\overrightarrow{a) }=0 (2) \\ \mathrm{From} (1) \mathrm{and} (2) \\ \overrightarrow{a.}\overrightarrow{c} =\overrightarrow{a}.\overrightarrow{b} \\ \mathrm{And} \overrightarrow{c}.\overrightarrow{b} =\overrightarrow{c}.\overrightarrow{a} \\ \mathrm{Hence} \overrightarrow{c}.\overrightarrow{b} =\overrightarrow{a}.\overrightarrow{b} \\ \mathrm{Or} \overrightarrow{b}.(\overrightarrow{c}-\overrightarrow{a}) =0 \\ \mathrm{Hence} HB \mathrm{is} \mathrm{perpendicular} \mathrm{to} AC. \end{gathered}$$