17. Determine whether or not the three given vectors are coplanar. $$ \begin{array}{l} ext { (a) } u=\left(\begin{array}{c}2 \ -1 \ 2\end{array} ight), v=\left(\begin{array}{c}4 \ 1 \ -1\end{array} ight), \boldsymbol{w}=\left(\begin{array}{c}6 \ -3 \ 1\end{array} ight) \ ext { (b) } u=\left(\begin{array}{c}4 \ -2 \ -1\end{array} ight), v=\left(\begin{array}{c}9 \ -6 \ -1\end{array} ight), \boldsymbol{w}=\left(\begin{array}{c}6 \ -6 \ 1\end{array} ight)\end{array} $$

Question

17. Determine whether or not the three given vectors are coplanar. $$ \begin{array}{l}	ext { (a) } u=\left(\begin{array}{c}2 \ -1 \ 2\end{array}ight), v=\left(\begin{array}{c}4 \ 1 \ -1\end{array}ight), \boldsymbol{w}=\left(\begin{array}{c}6 \ -3 \ 1\end{array}ight) \ 	ext { (b) } u=\left(\begin{array}{c}4 \ -2 \ -1\end{array}ight), v=\left(\begin{array}{c}9 \ -6 \ -1\end{array}ight), \boldsymbol{w}=\left(\begin{array}{c}6 \ -6 \ 1\end{array}ight)\end{array} $$

UpStudy AI · Accepted Answer

Let 
$$
u = \begin{pmatrix} u_1 \ u_2 \ u_3 \end{pmatrix}, \quad v = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix}, \quad w = \begin{pmatrix} w_1 \ w_2 \ w_3 \end{pmatrix}.
$$
The three vectors are coplanar if and only if the triple scalar product 
$$
u \cdot (v \times w)
$$
is zero.

---

### (a)

The vectors are
$$
u = \begin{pmatrix} 2 \ -1 \ 2 \end{pmatrix}, \quad v = \begin{pmatrix} 4 \ 1 \ -1 \end{pmatrix}, \quad w = \begin{pmatrix} 6 \ -3 \ 1 \end{pmatrix}.
$$

**Step 1. Compute $$v \times w$$:**

Using the formula
$$
v \times w = \begin{pmatrix}
v_2 w_3 - v_3 w_2 \
v_3 w_1 - v_1 w_3 \
v_1 w_2 - v_2 w_1 
\end{pmatrix},
$$
we have

- First component:
  $$
  v_2 w_3 - v_3 w_2 = 1 \cdot 1 - (-1)(-3) = 1 - 3 = -2.
  $$

- Second component:
  $$
  v_3 w_1 - v_1 w_3 = (-1) \cdot 6 - 4 \cdot 1 = -6 - 4 = -10.
  $$

- Third component:
  $$
  v_1 w_2 - v_2 w_1 = 4 \cdot (-3) - 1 \cdot 6 = -12 - 6 = -18.
  $$

Thus,
$$
v \times w = \begin{pmatrix} -2 \ -10 \ -18 \end{pmatrix}.
$$

**Step 2. Compute $$u \cdot (v \times w)$$:**

$$
u \cdot (v \times w) = 2(-2) + (-1)(-10) + 2(-18) = -4 + 10 - 36 = -30.
$$

Since 
$$
u \cdot (v \times w) = -30 \neq 0,
$$
the vectors in part (a) are **not coplanar**.

---

### (b)

The vectors are
$$
u = \begin{pmatrix} 4 \ -2 \ -1 \end{pmatrix}, \quad v = \begin{pmatrix} 9 \ -6 \ -1 \end{pmatrix}, \quad w = \begin{pmatrix} 6 \ -6 \ 1 \end{pmatrix}.
$$

**Step 1. Compute $$v \times w$$:**

Again using
$$
v \times w = \begin{pmatrix}
v_2 w_3 - v_3 w_2 \
v_3 w_1 - v_1 w_3 \
v_1 w_2 - v_2 w_1 
\end{pmatrix},
$$
we calculate

- First component:
  $$
  v_2 w_3 - v_3 w_2 = (-6) \cdot 1 - (-1)(-6) = -6 - 6 = -12.
  $$

- Second component:
  $$
  v_3 w_1 - v_1 w_3 = (-1) \cdot 6 - 9 \cdot 1 = -6 - 9 = -15.
  $$

- Third component:
  $$
  v_1 w_2 - v_2 w_1 = 9 \cdot (-6) - (-6) \cdot 6 = -54 + 36 = -18.
  $$

Thus,
$$
v \times w = \begin{pmatrix} -12 \ -15 \ -18 \end{pmatrix}.
$$

**Step 2. Compute $$u \cdot (v \times w)$$:**

$$
u \cdot (v \times w) = 4(-12) + (-2)(-15) + (-1)(-18) = -48 + 30 + 18 = 0.
$$

Since 
$$
u \cdot (v \times w) = 0,
$$
the vectors in part (b) are **coplanar**.

---

### Final Answer

- Part (a): The vectors are **not coplanar**.
- Part (b): The vectors are **coplanar**.
- Part (a): The vectors are **not coplanar**. - Part (b): The vectors are **coplanar**.

Determine whether or not the three given vectors are coplanar.

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(a)

(b)

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