Let p^×q→×p^+p^⋅q→q→=x^2+y^2q→+14-4x-6yp^, where p^ and q→ are non-collinear vectors ( p^ is a unit vector) and x, y are scalars, then the value of…

Question

Let p^×q→×p^+p^⋅q→q→=x^2+y^2q→+14-4x-6yp^, where p^ and q→ are non-collinear vectors ( p^ is a unit vector) and x, y are scalars, then the value of x^2+y^2 is equal to

MARKS Content Team · Accepted Answer

Correct answer: (D) 13. p^×q→×p^+p^⋅q→q→=x^2+y^2q→+14-4x-6yp^ p^^2q→-p^⋅q→p^+p^⋅q→q→=x^2+y^2q→+14-4x-6yp^ Comparing both sides, we get, 14-4x-6y=-p^⋅q→ x^2+y^2=1+p^⋅q→ x^2+y^2-4x-6y+14=1 x^2+y^2-4x-6y+13=0 x-2^2+y-3^2=0⇒x=2, y=3

Let $(\hat{p} \times \vec{q}) \times \hat{p} + (\hat{p} \cdot \vec{q}) \vec{q} = (x^{2} + y^{2}) \vec{q} + (14 - 4 x - 6 y) \hat{p}$ , where $\hat{p}$ and $\vec{q}$ are non-collinear vectors ( $\hat{p}$ is a unit vector) and $x, y$ are scalars, then the value of $x^{2} + y^{2}$ is equal to

Solution

More Vector Algebra Questions

Let p^×q→×p^+p^⋅q→q→=x2+y2q→+14-4x-6yp^, where p^ and q→ are non-collinear vectors ( p^ is a unit vector) and x, y are scalars, then the value of x2+y2 is equal to

Solution

More Vector Algebra Questions

Let $(\hat{p} \times \vec{q}) \times \hat{p} + (\hat{p} \cdot \vec{q}) \vec{q} = (x^{2} + y^{2}) \vec{q} + (14 - 4 x - 6 y) \hat{p}$ , where $\hat{p}$ and $\vec{q}$ are non-collinear vectors ( $\hat{p}$ is a unit vector) and $x, y$ are scalars, then the value of $x^{2} + y^{2}$ is equal to