The position vector of P- O-P- - xi - yj - zk where x-y-z - N and a - i - j - k- If O-P-a - 18 then the number of possible positions of P is

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Standard XII

Mathematics

Parametric Form of Tangent: Ellipse

The position ...

Question

The position vector of P, $¯ O P = x i + y j + z k$ where $x, y, z \in N$ and a $= i + j + k$ . If $¯ O P . a =$ 18 then the number of possible positions of P is

272

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306

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153

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136

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Solution

The correct option is C 136
$\to O P . \to a$
$= (x i + y j + z k) . (i + j + k)$
$= x + y + z$
$= 18$
Therefore
$x + y + z = 18$
Then the number of possible integral solutions
$= \frac{(n - 1) . (n - 2)}{2}$

$= \frac{(18 - 1) . (18 - 2)}{2}$

$= \frac{17.16}{2}$
$= \frac{272}{2}$
$= 136$ .

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The position vector of P, ¯OP=xi+yj+zk where x,y,z∈N and a =i+j+k. If ¯OP.a= 18 then the number of possible positions of P is

The correct option is C 136→OP.→a=(xi+yj+zk).(i+j+k)=x+y+z=18Thereforex+y+z=18Then the number of possible integral solutions =(n−1).(n−2)2=(18−1).(18−2)2=17.162=2722=136.

The position vector of P, $¯ O P = x i + y j + z k$ where $x, y, z \in N$ and a $= i + j + k$ . If $¯ O P . a =$ 18 then the number of possible positions of P is

The correct option is C 136
$\to O P . \to a$
$= (x i + y j + z k) . (i + j + k)$
$= x + y + z$
$= 18$
Therefore
$x + y + z = 18$
Then the number of possible integral solutions
$= \frac{(n - 1) . (n - 2)}{2}$

$= \frac{(18 - 1) . (18 - 2)}{2}$

$= \frac{17.16}{2}$
$= \frac{272}{2}$
$= 136$ .