The position vector of a point P is r-x-y-zk- where x-y-z - N and a-k- If r-a-12- then number of possible position of P is

The correct option is B 55 Let r=xi+yj+zk Thus, r.a=x+y+z Hence, x+y+z=12 = gt; x+y+z=12 This is thus equivalent to distributing ...

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Standard XII

Mathematics

Properties of Modulus

The position ...

Question

The position vector of a point $P$ is $¯ r = x^i + y^j + z^k$ where $x, y, z ϵ N$ and $¯ a =^i +^j +^k$ . If $¯ r \cdot ¯ a = 12$ , then number of possible position of P is

45

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Solution

The correct option is B $55$
Let $r = x i + y j + z k$
Thus, $r . a = x + y + z$
Hence, $x + y + z = 12$
$=> x + y + z = 12$
This is thus equivalent to distributing $12$ identical sweets among $3$ boys.
Now, since x, y, z are positive integers, we give them 1 sweet each. Thus, we are left with $12 - 3 = 9$
Thus, we need to distribute $9$ sweets among $3$ boys where a boy getting no sweet is also allowed.
This can be done in $^{9 + 3 - 1} C_{3 - 1} =^{11} C_{2} = 55$ ways.
Hence, (b) is correct.

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The position vector of a point P is ¯r=x^i+y^j+z^k where x,y,zϵN and ¯a=^i+^j+^k. If ¯r⋅¯a=12, then number of possible position of P is

The position vector of a point $P$ is $¯ r = x^i + y^j + z^k$ where $x, y, z ϵ N$ and $¯ a =^i +^j +^k$ . If $¯ r \cdot ¯ a = 12$ , then number of possible position of P is