Question

p =2i^ + 3j -k and Q=2i-5j+2k .fine p+ q and 3p_2q
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Answer 1

$p+q= (4\hat{i}-2\hat{j}+\hat{k})$ and $3p-2q= (24\hat{j}-9\hat{k})$

Step-by-step Explanation:

This question uses the arithmetic concepts of simple vector algebra.

Given:

• Vector $p=2\hat{i} + 3\hat{j}-\hat{k}$
• Vector $q= 2\hat{i}-5\hat{j}+2\hat{k}$

To be found:

• Vector $p + q$
• Vector $3p- 2q$

Concepts used:

• $\hat{i}, \hat{j},\hat{k}$ represent unit vectors in the x, y and z axes. We can represent any 3D vector in terms of these.
• Addition rule of vectors: $(a_1\hat{i}+b_1\hat{j}+c_1\hat{k}) + (a_2\hat{i}+b_2\hat{j}+c_2\hat{k}) = (a_1+a_2)\hat{i}+(b_1+b_2)\hat{j}+(c_1+c_2)\hat{k}$  (we normally add each of the x, y and z components and write the final vector)
• Scalar Multiplication rule : $x*(a_1\hat{i}+b_1\hat{j}+c_1\hat{k}) = (xa_1\hat{i}+xb_1\hat{j}+xc_1\hat{k})$

We can add p and q like this,

$p+q=(2\hat{i}+3\hat{j}-\hat{k}) + (2\hat{i}-5\hat{j}+2\hat{k}) = (4\hat{i}-2\hat{j}+\hat{k})$

Also, we can multiply 3 to p and 2 to q (by using scalar multiplication rule):  

$3p=3(2\hat{i}+3\hat{j}-\hat{k})=(6\hat{i}+9\hat{j}-3\hat{k})$

$2q=3(2\hat{i}-5\hat{j}+2\hat{k})=(6\hat{i}-15\hat{j}+6\hat{k})$

Now we can evaluate $(3p-2q)$ by simple subtraction,

$3p-2q= (6-6)\hat{i}+(9+15)\hat{j}+(-3-6)\hat{k} = 24\hat{j}-9\hat{k}$

So, $p+q= (4\hat{i}-2\hat{j}+\hat{k})$ and $3p-2q= 24\hat{j}-9\hat{k}$