Answer :
To find the value of [tex]\( a \)[/tex] given that the distance between the points [tex]\( U = (2, -4, 1, a) \)[/tex] and [tex]\( V = (3, -1, a+10, -3) \)[/tex] is 6, we need to apply the distance formula for points in four-dimensional space.
The distance formula in four-dimensional space is:
[tex]\[ \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 + (w_1 - w_2)^2} \][/tex]
Given points [tex]\( U = (2, -4, 1, a) \)[/tex] and [tex]\( V = (3, -1, a+10, -3) \)[/tex], we plug these coordinates into the distance formula:
[tex]\[ \sqrt{(2 - 3)^2 + (-4 - (-1))^2 + (1 - (a + 10))^2 + (a - (-3))^2} = 6 \][/tex]
Simplify the expressions inside the square root:
[tex]\[ \sqrt{(2 - 3)^2 + (-4 + 1)^2 + (1 - a - 10)^2 + (a + 3)^2} = 6 \][/tex]
[tex]\[ \sqrt{(-1)^2 + (-3)^2 + (-9 - a)^2 + (a + 3)^2} = 6 \][/tex]
Evaluate the squares:
[tex]\[ \sqrt{1 + 9 + (a+9)^2 + (a+3)^2} = 6 \][/tex]
Expand the squares:
[tex]\[ (a+9)^2 = a^2 + 18a + 81 \][/tex]
[tex]\[ (a+3)^2 = a^2 + 6a + 9 \][/tex]
Now substitute these expansions back into the equation:
[tex]\[ \sqrt{1 + 9 + a^2 + 18a + 81 + a^2 + 6a + 9} = 6 \][/tex]
Combine like terms:
[tex]\[ \sqrt{2a^2 + 24a + 100} = 6 \][/tex]
Square both sides to eliminate the square root:
[tex]\[ 2a^2 + 24a + 100 = 36 \][/tex]
Subtract 36 from both sides:
[tex]\[ 2a^2 + 24a + 64 = 0 \][/tex]
Divide the entire equation by 2 to simplify:
[tex]\[ a^2 + 12a + 32 = 0 \][/tex]
Now, solve this quadratic equation using the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = 32 \)[/tex]:
[tex]\[ a = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} \][/tex]
[tex]\[ a = \frac{-12 \pm \sqrt{144 - 128}}{2} \][/tex]
[tex]\[ a = \frac{-12 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ a = \frac{-12 \pm 4}{2} \][/tex]
This gives us two solutions:
[tex]\[ a = \frac{-12 + 4}{2} = \frac{-8}{2} = -4 \][/tex]
[tex]\[ a = \frac{-12 - 4}{2} = \frac{-16}{2} = -8 \][/tex]
Therefore, the values of [tex]\( a \)[/tex] that satisfy the given conditions are [tex]\( \boxed{-4 \text{ and } -8} \)[/tex].
The distance formula in four-dimensional space is:
[tex]\[ \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 + (w_1 - w_2)^2} \][/tex]
Given points [tex]\( U = (2, -4, 1, a) \)[/tex] and [tex]\( V = (3, -1, a+10, -3) \)[/tex], we plug these coordinates into the distance formula:
[tex]\[ \sqrt{(2 - 3)^2 + (-4 - (-1))^2 + (1 - (a + 10))^2 + (a - (-3))^2} = 6 \][/tex]
Simplify the expressions inside the square root:
[tex]\[ \sqrt{(2 - 3)^2 + (-4 + 1)^2 + (1 - a - 10)^2 + (a + 3)^2} = 6 \][/tex]
[tex]\[ \sqrt{(-1)^2 + (-3)^2 + (-9 - a)^2 + (a + 3)^2} = 6 \][/tex]
Evaluate the squares:
[tex]\[ \sqrt{1 + 9 + (a+9)^2 + (a+3)^2} = 6 \][/tex]
Expand the squares:
[tex]\[ (a+9)^2 = a^2 + 18a + 81 \][/tex]
[tex]\[ (a+3)^2 = a^2 + 6a + 9 \][/tex]
Now substitute these expansions back into the equation:
[tex]\[ \sqrt{1 + 9 + a^2 + 18a + 81 + a^2 + 6a + 9} = 6 \][/tex]
Combine like terms:
[tex]\[ \sqrt{2a^2 + 24a + 100} = 6 \][/tex]
Square both sides to eliminate the square root:
[tex]\[ 2a^2 + 24a + 100 = 36 \][/tex]
Subtract 36 from both sides:
[tex]\[ 2a^2 + 24a + 64 = 0 \][/tex]
Divide the entire equation by 2 to simplify:
[tex]\[ a^2 + 12a + 32 = 0 \][/tex]
Now, solve this quadratic equation using the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = 32 \)[/tex]:
[tex]\[ a = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} \][/tex]
[tex]\[ a = \frac{-12 \pm \sqrt{144 - 128}}{2} \][/tex]
[tex]\[ a = \frac{-12 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ a = \frac{-12 \pm 4}{2} \][/tex]
This gives us two solutions:
[tex]\[ a = \frac{-12 + 4}{2} = \frac{-8}{2} = -4 \][/tex]
[tex]\[ a = \frac{-12 - 4}{2} = \frac{-16}{2} = -8 \][/tex]
Therefore, the values of [tex]\( a \)[/tex] that satisfy the given conditions are [tex]\( \boxed{-4 \text{ and } -8} \)[/tex].