You are given vectors A⃗ = 4.8 i^− 6.7 j^ and B⃗ = - 3.4 i^+ 7.5 j^. A third vector C⃗ lies in the xy-plane. Vector C⃗ is perpendicular to vector A⃗ and the scalar product of C⃗ with B⃗ is 17.0.
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To find vector C⃗, we need to find a vector that is perpendicular to vector A⃗.
When two vectors are perpendicular, their dot product is zero. So, we can use the dot product to find a vector perpendicular to A⃗.
The dot product of two vectors A⃗ and B⃗ is calculated as:
A⃗ · B⃗ = |A| |B| cosθ
Where |A| and |B| are the magnitudes of vectors A⃗ and B⃗, and θ is the angle between the two vectors.
Since we want vector C⃗ to be perpendicular to vector A⃗, the dot product between C⃗ and A⃗ should be zero:
C⃗ · A⃗ = 0
Let's calculate the dot product of C⃗ and A⃗:
C⃗ · A⃗ = (C_x i^ + C_y j^) · (4.8 i^ - 6.7 j^)
Using the dot product formula, we can expand it as follows:
C_x * 4.8 + C_y * (-6.7) = 0
Now, we have one equation with two unknowns (C_x and C_y). This means we need more information to find the values of C_x and C_y.
The second piece of information given is the scalar product of C⃗ and B⃗, which is 17.0. The scalar product is calculated as:
C⃗ · B⃗ = |C| |B| cosθ
Since we know that C⃗ is perpendicular to A⃗, the angle between C⃗ and B⃗ is 90 degrees, and cos90° = 0.
So, the scalar product between C⃗ and B⃗ simplifies to:
C⃗ · B⃗ = |C| |B| cos90° = 0
From the given information, we have C⃗ · B⃗ = 17.0. But the scalar product of C⃗ and B⃗ should be zero, so there is a contradiction. It seems that there is no solution that satisfies both the perpendicularity condition and the scalar product condition.
Therefore, there is no vector C⃗ that is both perpendicular to A⃗ and has a scalar product of 17.0 with B⃗.