JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 1)

Then the number of such points for which $$x^2 + {y^2} + {z^2} \le 100$$ is
Answer
7

Explanation

To solve this problem, we need to find the integer points $$\left( {x,\,y,\,z} \right)$$ that satisfy the given system of homogeneous equations:

$$\matrix{ {3x - y - z = 0} \cr { - 3x + z = 0} \cr { - 3x + 2y + z = 0} \cr }$$

Firstly, let’s solve for $$z$$ in terms of $$x$$ from the second equation:

$$ - 3x + z = 0 \Rightarrow z = 3x $$

Next, substitute $$z = 3x$$ into the first equation:

$$ 3x - y - 3x = 0 \Rightarrow -y = 0 \Rightarrow y = 0 $$

With $$y = 0$$ and $$z = 3x$$, the third equation also should be satisfied. Let's substitute $$y$$ and $$z$$ back into the third equation to verify:

$$ - 3x + 2y + z = 0 \Rightarrow - 3x + 2(0) + 3x = 0 $$

This equation holds true, confirming that the solutions for $$y$$ and $$z$$ remain consistent. Therefore, the points that satisfy the given system are of the form:

$$\left( x,\,0,\,3x \right)$$

Additionally, we need $$x^2 + y^2 + z^2 \le 100$$. Substituting $$y = 0$$ and $$z = 3x$$, we get:

$$ x^2 + 0^2 + (3x)^2 \le 100 $$

This further simplifies to:

$$ x^2 + 9x^2 \le 100 $$

$$ 10x^2 \le 100 $$

$$ x^2 \le 10 $$

Hence, $$ -\sqrt{10} \le x \le \sqrt{10} $$

Since $$x$$ must be an integer, we evaluate acceptable values for $$x$$:

$$x \in \{-3,\,-2,\,-1,\,0,\,1,\,2,\,3\}$$

For each of these values, let’s determine the corresponding points $$\left( x,\,0,\,3x \right)$$:

  • $$( -3,\,0,\,-9 )$$
  • $$( -2,\,0,\,-6 )$$
  • $$( -1,\,0,\,-3 )$$
  • $$( 0,\,0,\,0 )$$
  • $$( 1,\,0,\,3 )$$
  • $$( 2,\,0,\,6 )$$
  • $$( 3,\,0,\,9 )$$

Thus, there are a total of 7 such points.

Therefore, the number of integer-coordinate points $$\left( x,\,y,\,z \right)$$ satisfying the given system of equations and the condition $$x^2 + y^2 + z^2 \le 100$$ is 7.

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