Given vectors $s = -i + 2j$, $t = 3i - j$, and $r = 2i + 5j$, we need to find expressions for: (a) $s+t$ (b) $r-s$ (c) $2t+r$ and then determine which of the options (A) $3s - 2r$ and (B) $\frac{1}{2}(2t - r)$ are equivalent to one of (a), (b), or (c).

1. Problem Description

Given vectors $s = -i + 2j$, $t = 3i - j$, and $r = 2i + 5j$, we need to find expressions for:

(a) $s+t$

(b) $r-s$

(c) $2t+r$

and then determine which of the options (A) $3s - 2r$ and (B) $\frac{1}{2}(2t - r)$ are equivalent to one of (a), (b), or (c).

2. Solution Steps

First, let's compute (a), (b), and (c):

(a) $s+t = (-i+2j) + (3i-j) = (-1+3)i + (2-1)j = 2i + j$

(b) $r-s = (2i+5j) - (-i+2j) = (2-(-1))i + (5-2)j = 3i + 3j$

(c) $2t+r = 2(3i-j) + (2i+5j) = (6i - 2j) + (2i + 5j) = (6+2)i + (-2+5)j = 8i + 3j$

Now, let's compute (A) and (B):

(A) $3s - 2r = 3(-i+2j) - 2(2i+5j) = (-3i+6j) - (4i+10j) = (-3-4)i + (6-10)j = -7i - 4j$

(B) $\frac{1}{2}(2t - r) = \frac{1}{2}(2(3i-j) - (2i+5j)) = \frac{1}{2}((6i-2j) - (2i+5j)) = \frac{1}{2}((6-2)i + (-2-5)j) = \frac{1}{2}(4i - 7j) = 2i - \frac{7}{2}j$

Now, let's compare:

$s+t = 2i+j$

$r-s = 3i+3j$

$2t+r = 8i+3j$

$3s-2r = -7i-4j$

$\frac{1}{2}(2t-r) = 2i - \frac{7}{2}j$

None of the options (A) and (B) match any of the results (a), (b), or (c). There might be a typo in the original question, or a calculation error. Let us suppose there is a typo in the question and option A is equal to r - s, so let us solve the question for what values of s and r are $3s-2r = r-s$

$4s = 3r$

$s = (3/4)r$

Let us suppose there is a typo in the question and option B is equal to r - s, so let us solve the question for what values of t and r are $(1/2)(2t-r) = r-s$

$2t-r = 2r - 2s$

$2t = 3r - 2s$

$t = (3/2)r - s$

Let us continue comparing. If the question meant $r-t$, then

$r-t = (2i+5j) - (3i-j) = (2-3)i + (5-(-1))j = -i+6j$.

Let us look for another possibility. Since $s = -i + 2j$, $3i = -3s + 6j$, but this doesnt work.

(A) $3s - 2r = -7i-4j$ is not $r-s = 3i+3j$

(B) $\frac{1}{2}(2t-r) = 2i - \frac{7}{2}j$ is not $r-s = 3i+3j$

Let's compare (A) with $3s-2r$. It is not $s+t$, $r-s$, or $2t+r$.

Let's compare (B) with $\frac{1}{2}(2t-r)$. It is not $s+t$, $r-s$, or $2t+r$.

It may be that there are some transcription errors.

3. Final Answer

None of the options (A) $3s-2r$ and (B) $\frac{1}{2}(2t-r)$ are equal to (a) $s+t$, (b) $r-s$, or (c) $2t+r$.

(a) $s+t = 2i+j$

(b) $r-s = 3i+3j$

(c) $2t+r = 8i+3j$

(A) $3s-2r = -7i-4j$

(B) $\frac{1}{2}(2t-r) = 2i - \frac{7}{2}j$

Final Answer: None