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The projection of the vector 2ari - 3arj on the vector 3ari + 4arj is

Mathematics

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Answer

E. 65\text{E. } \frac{6}{5}

Step-by-step solution

5 steps
  1. Identify the two vectors.

    Let a⃗\vec{a} be the vector 2i^−3k^2\hat{i} - 3\hat{k}.
    Let b⃗\vec{b} be the vector 3i^+4j^3\hat{i} + 4\hat{j}.

  2. Calculate the dot product of a⃗\vec{a} and b⃗\vec{b}.

    The dot product a⃗⋅b⃗\vec{a} \cdot \vec{b} is given by:

    a⃗⋅b⃗=(2)(3)+(0)(4)+(−3)(0)\vec{a} \cdot \vec{b} = (2)(3) + (0)(4) + (-3)(0) a⃗⋅b⃗=6+0+0\vec{a} \cdot \vec{b} = 6 + 0 + 0 a⃗⋅b⃗=6\vec{a} \cdot \vec{b} = 6
  3. Calculate the magnitude of vector b⃗\vec{b}.

    The magnitude of b⃗\vec{b}, denoted as ∣b⃗∣|\vec{b}|, is given by:

    ∣b⃗∣=32+42+02|\vec{b}| = \sqrt{3^2 + 4^2 + 0^2} ∣b⃗∣=9+16|\vec{b}| = \sqrt{9 + 16} ∣b⃗∣=25|\vec{b}| = \sqrt{25} ∣b⃗∣=5|\vec{b}| = 5
  4. Calculate the projection of a⃗\vec{a} on b⃗\vec{b}.

    The projection of vector a⃗\vec{a} on vector b⃗\vec{b} is given by the formula:

    projb⃗a⃗=a⃗⋅b⃗∣b⃗∣\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

    Substitute the values calculated in Step 2 and Step 3:

    projb⃗a⃗=65\text{proj}_{\vec{b}} \vec{a} = \frac{6}{5}
  5. Compare the result with the given options.

    The calculated projection is 65\frac{6}{5}, which corresponds to option E.

    The final answer is E. 65\text{E. } \frac{6}{5}.

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