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Question: The plane $\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1$ intersect x-axis, y-axis, z-axis at A, B, C ...

The plane x1+y2+z3=1\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1 intersect x-axis, y-axis, z-axis at A, B, C respectively. If the distance between origin and othrocenter of △ABC\triangle ABC is equal to k then value of 7k is equal to

Answer

6

Explanation

Solution

Solution:

  1. Find the intercepts:
    The plane is given by

    x1+y2+z3=1.\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1.

    The intercepts are:

    A=(1,0,0),B=(0,2,0),C=(0,0,3).A=(1,0,0),\quad B=(0,2,0),\quad C=(0,0,3).

  2. Express the orthocenter H as (h₁, h₂, h₃):
    The altitude from A is perpendicular to BC.

    BC⃗=C−B=(0,0,3)−(0,2,0)=(0,−2,3).\vec{BC} = C-B = (0,0,3)-(0,2,0) = (0,-2,3).

    With H = (h₁, h₂, h₃), the condition for the altitude from A:

    (h1−1, h2−0, h3−0)⋅(0,−2,3)=0⇒−2h2+3h3=0.(h₁-1,\, h₂-0,\, h₃-0)\cdot (0,-2,3)=0\quad\Rightarrow\quad -2h₂+3h₃=0.

    Hence,

    h2=32h3.(i)h₂=\frac{3}{2}h₃.\quad\text{(i)}

    Similarly, the altitude from B is perpendicular to AC.

    AC⃗=C−A=(0−1, 0−0, 3−0)=(−1,0,3).\vec{AC} = C-A = (0-1,\, 0-0,\, 3-0) = (-1,0,3).

    For B=(0,2,0), the altitude condition gives:

    (h1−0, h2−2, h3−0)⋅(−1,0,3)=0⇒−h1+3h3=0.(h₁-0,\, h₂-2,\, h₃-0)\cdot (-1,0,3)=0\quad\Rightarrow\quad -h₁+3h₃=0.

    Thus,

    h1=3h3.(ii)h₁=3h₃.\quad\text{(ii)}

  3. Use the fact that H lies on the plane:
    Substitute H = (3t, (3/2)t, t) (taking h₃=t) into the plane equation:

    3t1+(3/2)t2+t3=1.\frac{3t}{1}+\frac{(3/2)t}{2}+\frac{t}{3}=1.

    Simplify:

    3t+3t4+t3=1.3t + \frac{3t}{4} + \frac{t}{3} = 1.

    Multiply through by 12 (LCM of 1,4,3):

    36t+9t+4t=12⟹49t=12.36t + 9t + 4t = 12 \quad\Longrightarrow\quad 49t = 12.

    So,

    t=1249.t=\frac{12}{49}.

    Therefore, the orthocenter is:

    H=(3t,  32t,  t)=(3649, 1849, 1249).H=\left(3t,\; \frac{3}{2}t,\; t\right)=\left(\frac{36}{49},\, \frac{18}{49},\, \frac{12}{49}\right).

  4. Calculate the distance from the origin to H (k):

    k=(3649)2+(1849)2+(1249)2.k = \sqrt{\left(\frac{36}{49}\right)^2+\left(\frac{18}{49}\right)^2+\left(\frac{12}{49}\right)^2}.

    Compute:

    (3649)2=12962401,(1849)2=3242401,(1249)2=1442401.\left(\frac{36}{49}\right)^2 = \frac{1296}{2401},\quad \left(\frac{18}{49}\right)^2 = \frac{324}{2401},\quad \left(\frac{12}{49}\right)^2 = \frac{144}{2401}.

    Summing, we get:

    k=1296+324+1442401=17642401=4249=67.k = \sqrt{\frac{1296+324+144}{2401}} = \sqrt{\frac{1764}{2401}} = \frac{42}{49} = \frac{6}{7}.

  5. Find 7k:

    7k=7⋅67=6.7k = 7\cdot\frac{6}{7} = 6.

Summary:

  • Explanation: Find intercepts, express orthocenter H in terms of parameter t using perpendicularity conditions from vertices A and B, substitute H into the plane equation to find t, compute H, then the distance from O to H, and finally multiply by 7.

  • Answer to the question: 6