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Vector Algebra question

2015 · Shift 2 · Q23
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Vector Algebra question

2015 · Shift 2 · Q23

JEE AdvancedMathematicsVector AlgebraNumerical+4 / −1
Suppose that p→,q→\overrightarrow p ,\overrightarrow qp​,q​ and r→\overrightarrow rr are three non-coplanar vectors in R3{R^3}R3. Let the components of a vector s→\overrightarrow ss along p→,q→\overrightarrow p ,\overrightarrow qp​,q​ and r→\overrightarrow rr be 4,34, 34,3 and 5,5,5, respectively. If the components of this vector s→\overrightarrow ss along (−p→+q→+r→),(p→−q→+r→)\left( { - \overrightarrow p + \overrightarrow q + \overrightarrow r } \right),\left( {\overrightarrow p - \overrightarrow q + \overrightarrow r } \right)(−p​+q​+r),(p​−q​+r) and (−p→−q→+r→)\left( { - \overrightarrow p - \overrightarrow q + \overrightarrow r } \right)(−p​−q​+r) are x,yx, yx,y and z,z,z, respectively, then the value of 2x+y+z2x+y+z2x+y+z is
Numerical answer
View written solutionFree

Correct answer: 9

  1. Since the components of s⃗\vec ss along p⃗,q⃗,r⃗\vec p, \vec q, \vec rp​,q​,r are 4,3,54,3,54,3,5, we can write
s⃗=4p⃗+3q⃗+5r⃗.\vec s = 4\vec p + 3\vec q + 5\vec r.s=4p​+3q​+5r.
  1. Now we are told that the components of s⃗\vec ss along
(−p⃗+q⃗+r⃗),(p⃗−q⃗+r⃗),(−p⃗−q⃗+r⃗)(-\vec p+\vec q+\vec r),\quad (\vec p-\vec q+\vec r),\quad (-\vec p-\vec q+\vec r)(−p​+q​+r),(p​−q​+r),(−p​−q​+r)

are x,y,zx,y,zx,y,z respectively. Hence,

s⃗=x(−p⃗+q⃗+r⃗)+y(p⃗−q⃗+r⃗)+z(−p⃗−q⃗+r⃗).\vec s = x(-\vec p+\vec q+\vec r)+y(\vec p-\vec q+\vec r)+z(-\vec p-\vec q+\vec r).s=x(−p​+q​+r)+y(p​−q​+r)+z(−p​−q​+r).
  1. Expand the right-hand side:
s⃗=(−x+y−z)p⃗+(x−y−z)q⃗+(x+y+z)r⃗.\vec s = (-x+y-z)\vec p + (x-y-z)\vec q + (x+y+z)\vec r.s=(−x+y−z)p​+(x−y−z)q​+(x+y+z)r.
  1. Compare coefficients with
s⃗=4p⃗+3q⃗+5r⃗.\vec s = 4\vec p+3\vec q+5\vec r.s=4p​+3q​+5r.

So we get the system:

−x+y−z=4...(1)-x+y-z=4 \quad ...(1)−x+y−z=4...(1) x−y−z=3...(2)x-y-z=3 \quad ...(2)x−y−z=3...(2) x+y+z=5...(3)x+y+z=5 \quad ...(3)x+y+z=5...(3)
  1. Solve for x,y,zx,y,zx,y,z.

Add (1) and (2):

(−x+y−z)+(x−y−z)=4+3(-x+y-z)+(x-y-z)=4+3(−x+y−z)+(x−y−z)=4+3 −2z=7  ⟹  z=−72.-2z=7 \implies z=-\frac{7}{2}.−2z=7⟹z=−27​.

From (3):

x+y−72=5x+y-\frac{7}{2}=5x+y−27​=5 x+y=172....(4)x+y=\frac{17}{2}. \quad ...(4)x+y=217​....(4)

From (2):

x−y−(−72)=3x-y-\left(-\frac{7}{2}\right)=3x−y−(−27​)=3 x−y+72=3x-y+\frac{7}{2}=3x−y+27​=3 x−y=−12....(5)x-y=-\frac{1}{2}. \quad ...(5)x−y=−21​....(5)

Add (4) and (5):

2x=8  ⟹  x=4.2x=8 \implies x=4.2x=8⟹x=4.

Then

y=172−4=92.y=\frac{17}{2}-4=\frac{9}{2}.y=217​−4=29​.

So,

x=4,y=92,z=−72.x=4,\quad y=\frac{9}{2},\quad z=-\frac{7}{2}.x=4,y=29​,z=−27​.
  1. Now compute:
2x+y+z=2(4)+92−72=8+1=9.2x+y+z = 2(4)+\frac{9}{2}-\frac{7}{2} = 8+1=9.2x+y+z=2(4)+29​−27​=8+1=9.

Therefore, the required value is

9.\boxed{9}.9​.
  1. Comparison with stored correct answer:
  • Derived answer = 999
  • Stored correct answer = 999

They agree.

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