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Chapter 14: Problem 6

Find the modulus of the vector $$ p=2 i-j+5 k $$

Short Answer

Expert verified
Answer: The modulus of the vector \(p\) is \(\sqrt{30}\).

Step by step solution

01

Identify the Components of the Vector

Given the vector \(p = 2i - j + 5k\), we can see that the components of the vector are: $$ x = 2, \quad y = -1, \quad z = 5 $$
02

Apply the Modulus Formula

Use the formula mentioned above to find the modulus of the vector: $$ ||p|| = \sqrt{x^2 + y^2 + z^2} $$
03

Substitute the Components

Now, we will substitute the components we identified in the previous step, into the formula: $$ ||p|| = \sqrt{(2)^2 + (-1)^2 + (5)^2} $$
04

Calculate the Modulus

Calculate the sum of the squares and take the square root of the result to find the modulus of the vector: $$ ||p|| = \sqrt{4 + 1 + 25} = \sqrt{30} $$
05

Final Answer

The modulus of the given vector is: $$ ||p|| = \sqrt{30} $$

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Most popular questions from this chapter

Two unit vectors are parallel. What can you deduce about their scalar product?

On a diagram show the arbitrary vectors \(p\) and q. Then show the following: (a) \(p+q\) (d) \(4 q\) (e) \(-2 q\) (c) \(q-p\).

Explain the distinction between a position vector and a more general or free vector.

Find the work done by a force of magnitude 10 newtons acting in the direction of the vector \(3 \boldsymbol{i}+\boldsymbol{j}+8 \boldsymbol{k}\) if it moves a particle from the point \((1,1,1)\) to the point \((3,1,2)\).

If the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) is equal to zero, then (i) \(a=0\), or \(b=0\), or \(c=0\) or (ii) two of the vectors are parallel, or (iii) the three vectors lie in the same plane (they are said to be coplanar). Show that the vectors $$ 2 \boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}-4 \boldsymbol{j}+5 \boldsymbol{k}, \boldsymbol{i}+2 \boldsymbol{j}-3 \boldsymbol{k} $$ are coplanar.
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