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Displacement d1→is in the yz plane 63.0°from the positive direction of the y axis, has a positive z component, and has a magnitude of 4.50 m. Displacement is in the xz plane 30.0°from the positive direction of the x axis, has a positive z component, and has magnitude 1.40 m. What are (a)d→1.d→2, (b) role="math" localid="1656999023128" d→1×d→2, and (c) the angle between d→1and d→2?

Short Answer

Expert verified

(a) The dot product is 2.81.

(b) Cross product is 1.43iÁåœ+4.86jÁåœ-2.48k.

(c) The direction is 63.5°.

Step by step solution

01

Step 1: Given

d1=4.50cos63jÁåœ+sin63kÁåœ=2.04jÁåœ+4.01kÁåœd2=1.40cos30iÁåœ+sin30kÁåœ=1.21iÁåœ+0.7kÁåœ

02

Determining the concept

Here, use rule of vector product, dot product and cross product. Dot product of two vectors is scalar quantity and cross product of two vectors is vector quantity.

Formula is as follows:

cosθ=d1→.d2→d1→d2→

Where,

role="math" localid="1656999888380" d1→,d2→Are displacement vectors.

03

(a) Determining d1→.d2→

Using the formula for the dot product,

d1→.d2→=2.04jÁåœ+4.01kÁåœ.1.21iÁåœ+0.7kÁåœ)=2.81

Thus,the dot product is 2.81.

04

(b) Determining d1→×d2→

Using the formula for the cross product,

d1→×d2→=2.04jÁåœ+4.0kÁåœÃ—1.21iÁåœ+0.7kÁåœ=1.43iÁåœ+4.86jÁåœ-2.48kÁåœ

Thus, cross product is 1.43iÁåœ+4.86jÁåœ-2.48kÁåœ.

05

(c) Determining anangle

Magnitude ofd1 is as follow:

d1=2.042+4.012=4.50

Magnitude of is as follow:

d2=1.212+0.72=1.40

Therefore, the direction is given as,

cosθ=d1→.d2→d1→d2→=2.814.5×1.40θ=63.5°

Hence, thedirection is 63.5°.

Therefore, from the given information, vectors in the unit vector notation form can be written. Using the formula for the dot product and cross product, the answers can be found. It is also possible to find the angle between them using the property of dot product.

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