We read your photo as:

Given the vectors m = j + 2j and n = -3j + k. find: the angle between m and n, the projection of m on the n.

Mathematics

ScanSolve AI Answer

Answers

2 parts
a

arccos⁡(210)≈81.87∘\mathbf{\arccos\left(\frac{\sqrt{2}}{10}\right) \approx 81.87^\circ}

b

310i⃗+110j⃗\mathbf{\frac{3}{10}\vec{i} + \frac{1}{10}\vec{j}}

Step-by-step solution

5 steps

Here are the steps to solve the problem:

Given vectors are m⃗=j⃗+2k⃗\vec{m} = \vec{j} + 2\vec{k} and n⃗=3i⃗+j⃗\vec{n} = 3\vec{i} + \vec{j}.
First, write the vectors in component form:

m⃗=(012)andn⃗=(310)\vec{m} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \quad \text{and} \quad \vec{n} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}

Part a· 3 steps

The angle between m⃗\vec{m} and n⃗\vec{n}

  1. Calculate the dot product of m⃗\vec{m} and n⃗\vec{n}.

    m⃗⋅n⃗=(0)(3)+(1)(1)+(2)(0)=0+1+0=1\vec{m} \cdot \vec{n} = (0)(3) + (1)(1) + (2)(0) = 0 + 1 + 0 = 1

  2. Calculate the magnitudes of m⃗\vec{m} and n⃗\vec{n}.

    ∣m⃗∣=02+12+22=0+1+4=5|\vec{m}| = \sqrt{0^2 + 1^2 + 2^2} = \sqrt{0 + 1 + 4} = \sqrt{5} ∣n⃗∣=32+12+02=9+1+0=10|\vec{n}| = \sqrt{3^2 + 1^2 + 0^2} = \sqrt{9 + 1 + 0} = \sqrt{10}

  3. Use the dot product formula for the angle θ\theta between two vectors

    cos⁡θ=m⃗⋅n⃗∣m⃗∣∣n⃗∣\cos \theta = \frac{\vec{m} \cdot \vec{n}}{|\vec{m}| |\vec{n}|}.

    cos⁡θ=15⋅10=150=152\cos \theta = \frac{1}{\sqrt{5} \cdot \sqrt{10}} = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}}

    To rationalize the denominator:

    cos⁡θ=152×22=210\cos \theta = \frac{1}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{10} θ=arccos⁡(210)\theta = \arccos\left(\frac{\sqrt{2}}{10}\right) θ≈arccos⁡(0.14142)≈81.87∘\theta \approx \arccos(0.14142) \approx 81.87^\circ

    The angle between m⃗\vec{m} and n⃗\vec{n} is arccos⁡(210)≈81.87∘\boxed{\mathbf{\arccos\left(\frac{\sqrt{2}}{10}\right) \approx 81.87^\circ}}.

Part b

310i⃗+110j⃗\mathbf{\frac{3}{10}\vec{i} + \frac{1}{10}\vec{j}}

The projection of m⃗\vec{m} on n⃗\vec{n}

  1. Use the formula for the vector projection of m⃗\vec{m} on n⃗\vec{n}

    projn⃗m⃗=m⃗⋅n⃗∣n⃗∣2n⃗\text{proj}_{\vec{n}} \vec{m} = \frac{\vec{m} \cdot \vec{n}}{|\vec{n}|^2} \vec{n}.
    We already have m⃗⋅n⃗=1\vec{m} \cdot \vec{n} = 1 and ∣n⃗∣=10|\vec{n}| = \sqrt{10}, so ∣n⃗∣2=10|\vec{n}|^2 = 10.

  2. Substitute the values into the projection formula.

    projn⃗m⃗=110(3i⃗+j⃗)\text{proj}_{\vec{n}} \vec{m} = \frac{1}{10} (3\vec{i} + \vec{j}) projn⃗m⃗=310i⃗+110j⃗\text{proj}_{\vec{n}} \vec{m} = \frac{3}{10}\vec{i} + \frac{1}{10}\vec{j}

    The projection of m⃗\vec{m} on n⃗\vec{n} is 310i⃗+110j⃗\boxed{\mathbf{\frac{3}{10}\vec{i} + \frac{1}{10}\vec{j}}}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QGiven the vector m = 1+2j and n = 3j, find (i) The angle between m and n, (ii) The projection of m…QFind the value of m for which the vectors −2mi + 2j − 2k and 3i + 4mj + 3k perpendicular to each…QGiven vectors P = 3i − 2j − 4k and Q = ni + 2j − k, find the value of n if P and Q are…QCalculate the scalar projection of vector a = 2i + 3j − 2k onto vector b = i + 2j − 3k.QFind the sine of the angle between the vectors a = 3i − 2j + k and b = i + 2j − 3k
Browse by topic:Linear Algebra
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp