JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 11)
If for some $$\alpha $$ and $$\beta $$ in R, the intersection of the
following three places
x + 4y – 2z = 1
x + 7y – 5z = b
x + 5y + $$\alpha $$z = 5
is a line in R3, then $$\alpha $$ + $$\beta $$ is equal to :
x + 4y – 2z = 1
x + 7y – 5z = b
x + 5y + $$\alpha $$z = 5
is a line in R3, then $$\alpha $$ + $$\beta $$ is equal to :
-10
0
10
2
Explanation
For planes to intersect on a line there should be infinite solution of the
given system of equations.
For infinite solutions
$$\Delta $$ = $$\left| {\matrix{ 1 & 4 & { - 2} \cr 1 & 7 & { - 5} \cr 1 & 5 & \alpha \cr } } \right|$$ = 0
$$ \Rightarrow $$ 1(7$$\alpha $$ + 25) – 4($$\alpha $$ + 5) – 2(5 – 7) = 0
$$ \Rightarrow $$ 7$$\alpha $$ + 25 – 4$$\alpha $$ – 20 + 4 = 0
$$ \Rightarrow $$ 3$$\alpha $$ + 9 = 0
$$ \Rightarrow $$ $$\alpha $$ = -3
Also $$\Delta $$z = 0
$$ \Rightarrow $$ $$\left| {\matrix{ 1 & 4 & 1 \cr 1 & 7 & \beta \cr 1 & 5 & 5 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 1(35 – 5$$\beta $$) – 4(5 – $$\beta $$) + 1(5 – 7) = 0
$$ \Rightarrow $$ 35 - 5$$\beta $$ - 20 + 4$$\beta $$ - 2 = 0
$$ \Rightarrow $$ $$\beta $$ = 13
$$ \therefore $$ $$\alpha $$ + $$\beta $$ = -3 + 13 = 10
For infinite solutions
$$\Delta $$ = $$\left| {\matrix{ 1 & 4 & { - 2} \cr 1 & 7 & { - 5} \cr 1 & 5 & \alpha \cr } } \right|$$ = 0
$$ \Rightarrow $$ 1(7$$\alpha $$ + 25) – 4($$\alpha $$ + 5) – 2(5 – 7) = 0
$$ \Rightarrow $$ 7$$\alpha $$ + 25 – 4$$\alpha $$ – 20 + 4 = 0
$$ \Rightarrow $$ 3$$\alpha $$ + 9 = 0
$$ \Rightarrow $$ $$\alpha $$ = -3
Also $$\Delta $$z = 0
$$ \Rightarrow $$ $$\left| {\matrix{ 1 & 4 & 1 \cr 1 & 7 & \beta \cr 1 & 5 & 5 \cr } } \right|$$ = 0
$$ \Rightarrow $$ 1(35 – 5$$\beta $$) – 4(5 – $$\beta $$) + 1(5 – 7) = 0
$$ \Rightarrow $$ 35 - 5$$\beta $$ - 20 + 4$$\beta $$ - 2 = 0
$$ \Rightarrow $$ $$\beta $$ = 13
$$ \therefore $$ $$\alpha $$ + $$\beta $$ = -3 + 13 = 10
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