The position vector of a point P is r - x - y - z k- when x - y - z - N and a -k- - If r - a -10- the number of possible position of P isA- 72B- 36C- 66D- 9 C2

The correct option is B 36 r̅.a̅=10 ⇒ x + y + z =10, x, y, z ϵ N no. of solutions = ^10 1C_3 1=36 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

The position ...

Question

The position vector of a point P is $\to r = x^i + y^j + z^k$ , when $x, y, z ϵ$ N and $\to a =^i +^j +^k . I f \to r . \to a = 10$ , the number of possible position of P is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
36

$¯ r . ¯ a = 10$
$\Rightarrow x + y + z = 10, x, y, z ϵ N$
no. of solutions = $^{10 - 1} C_{3 - 1} = 36$

Suggest Corrections

Similar questions

Q. The position vector of a point

P

¯ 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

Q. The position vector of a point P is

\to r = x^i \to + y^j + z^k,

where

x ϵ N, y ϵ N, z ϵ N,

and

\to a = \to i + \to j + \to k \to

. If

\to r . \to a = 10

, the number of possible positions of P is

Q. If vectors

{\to a}_{1} = x^i -^j +^k

and

{\to a}_{2} =^i + y^j + 2^k

are collinear, then a possible unit vector parallel to the vector

x^i + y^j + z^k

is :

If the position vector of a point P is $\to r = x^i = y^j + z^k,$ where x, y, z $ϵ$ N and projection of $\to r$ on $\to a =^i +^j +^k i s \frac{10}{\sqrt{3}}$ then number of possible of P is also equal to

Q. If the vectors

x^i +^j +^k,^i + y^j +^k

and

^i +^j + z^k

are co-planar where

x \neq 1, y \neq 1, z \neq 1

, then prove that

\frac{1}{1 - x} + \frac{1}{1 - y} + \frac{1}{1 - z} = 1

Join BYJU'S Learning Program

The position vector of a point P is →r=x^i+y^j+z^k, when x,y,zϵ N and →a=^i+^j+^k.If→r.→a=10, the number of possible position of P is

The correct option is A 36 ¯r.¯a=10 ⇒x+y+z=10,x,y,zϵN no. of solutions = 10−1C3−1=36

The position vector of a point P is $\to r = x^i + y^j + z^k$ , when $x, y, z ϵ$ N and $\to a =^i +^j +^k . I f \to r . \to a = 10$ , the number of possible position of P is

The correct option is A
36

$¯ r . ¯ a = 10$
$\Rightarrow x + y + z = 10, x, y, z ϵ N$
no. of solutions = $^{10 - 1} C_{3 - 1} = 36$