Solved on Jan 22, 2024

Find the value of $k$ that makes the vectors $(2,-6,4)$ and $(1, k,-2)$ collinear.

STEP 1

Assumptions
1. Two vectors are collinear if one is a scalar multiple of the other.
2. The first vector is $(2, -6, 4)$ .
3. The second vector is $(1, k, -2)$ .
4. We need to find the value of $k$ that makes these two vectors collinear.

STEP 2

Let's set up the equation that represents the condition for two vectors to be collinear. If two vectors $\vec{a}$ and $\vec{b}$ are collinear, then there exists a scalar $\lambda$ such that $\vec{a} = \lambda \vec{b}$ .

STEP 3

Write the components of the vectors $\vec{a}$ and $\vec{b}$ , where $\vec{a} = (2, -6, 4)$ and $\vec{b} = (1, k, -2)$ , and equate them with the scalar multiple $\lambda$ .
$(2, -6, 4) = \lambda (1, k, -2)$

STEP 4

Equate each corresponding component of the vectors to find the value of $\lambda$ and $k$ .
$2 = \lambda \cdot 1$ $-6 = \lambda \cdot k$ $4 = \lambda \cdot (-2)$

STEP 5

Solve the first equation for $\lambda$ .
$\lambda = 2$

STEP 6

Substitute $\lambda = 2$ into the third equation to check for consistency.
$4 = 2 \cdot (-2)$

STEP 7

Verify if the third equation holds true.
$4 = -4$

STEP 8

Since $4 \neq -4$ , the third equation does not hold true, which implies that there is no such scalar $\lambda$ that can make the vectors collinear.

STEP 9

Conclude that there is no value of $k$ that will make the vectors $(2, -6, 4)$ and $(1, k, -2)$ collinear.
The correct answer is: A. There is no such $k$ .

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Find the value of $k$ that makes the vectors $(2,-6,4)$ and $(1, k,-2)$ collinear.

STEP 1

Assumptions
1. Two vectors are collinear if one is a scalar multiple of the other.
2. The first vector is $(2, -6, 4)$ .
3. The second vector is $(1, k, -2)$ .
4. We need to find the value of $k$ that makes these two vectors collinear.

STEP 2

Let's set up the equation that represents the condition for two vectors to be collinear. If two vectors $\vec{a}$ and $\vec{b}$ are collinear, then there exists a scalar $\lambda$ such that $\vec{a} = \lambda \vec{b}$ .

STEP 3

Write the components of the vectors $\vec{a}$ and $\vec{b}$ , where $\vec{a} = (2, -6, 4)$ and $\vec{b} = (1, k, -2)$ , and equate them with the scalar multiple $\lambda$ .
$(2, -6, 4) = \lambda (1, k, -2)$

STEP 4

Equate each corresponding component of the vectors to find the value of $\lambda$ and $k$ .
$2 = \lambda \cdot 1$ $-6 = \lambda \cdot k$ $4 = \lambda \cdot (-2)$

STEP 5

Solve the first equation for $\lambda$ .
$\lambda = 2$

STEP 6

Substitute $\lambda = 2$ into the third equation to check for consistency.
$4 = 2 \cdot (-2)$

STEP 7

Verify if the third equation holds true.
$4 = -4$

STEP 8

Since $4 \neq -4$ , the third equation does not hold true, which implies that there is no such scalar $\lambda$ that can make the vectors collinear.

STEP 9

Conclude that there is no value of $k$ that will make the vectors $(2, -6, 4)$ and $(1, k, -2)$ collinear.
The correct answer is: A. There is no such $k$ .

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Find the value of kkk that makes the vectors (2,−6,4)(2,-6,4)(2,−6,4) and (1,k,−2)(1, k,-2)(1,k,−2) collinear.

STEP 1

Assumptions1. Two vectors are collinear if one is a scalar multiple of the other.2. The first vector is (2,−6,4)(2, -6, 4)(2,−6,4).3. The second vector is (1,k,−2)(1, k, -2)(1,k,−2).4. We need to find the value of kkk that makes these two vectors collinear.

STEP 2

Let's set up the equation that represents the condition for two vectors to be collinear. If two vectors a⃗\vec{a}a and b⃗\vec{b}b are collinear, then there exists a scalar λ\lambdaλ such that a⃗=λb⃗\vec{a} = \lambda \vec{b}a=λb.

STEP 3

STEP 4

Equate each corresponding component of the vectors to find the value of λ\lambdaλ and kkk.2=λ⋅1 2 = \lambda \cdot 1 2=λ⋅1 −6=λ⋅k -6 = \lambda \cdot k −6=λ⋅k 4=λ⋅(−2) 4 = \lambda \cdot (-2) 4=λ⋅(−2)

STEP 5

Solve the first equation for λ\lambdaλ.λ=2 \lambda = 2 λ=2

STEP 6

Substitute λ=2\lambda = 2λ=2 into the third equation to check for consistency.4=2⋅(−2) 4 = 2 \cdot (-2) 4=2⋅(−2)

STEP 7

Verify if the third equation holds true.4=−4 4 = -4 4=−4

STEP 8

Since 4≠−44 \neq -44=−4, the third equation does not hold true, which implies that there is no such scalar λ\lambdaλ that can make the vectors collinear.

STEP 9

Conclude that there is no value of kkk that will make the vectors (2,−6,4)(2, -6, 4)(2,−6,4) and (1,k,−2)(1, k, -2)(1,k,−2) collinear.The correct answer is: A. There is no such kkk.

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Find the value of $k$ that makes the vectors $(2,-6,4)$ and $(1, k,-2)$ collinear.

Assumptions
1. Two vectors are collinear if one is a scalar multiple of the other.
2. The first vector is $(2, -6, 4)$ .
3. The second vector is $(1, k, -2)$ .
4. We need to find the value of $k$ that makes these two vectors collinear.

Let's set up the equation that represents the condition for two vectors to be collinear. If two vectors $\vec{a}$ and $\vec{b}$ are collinear, then there exists a scalar $\lambda$ such that $\vec{a} = \lambda \vec{b}$ .

Equate each corresponding component of the vectors to find the value of $\lambda$ and $k$ .
$2 = \lambda \cdot 1$ $-6 = \lambda \cdot k$ $4 = \lambda \cdot (-2)$

Solve the first equation for $\lambda$ .
$\lambda = 2$

Substitute $\lambda = 2$ into the third equation to check for consistency.
$4 = 2 \cdot (-2)$

Verify if the third equation holds true.
$4 = -4$

Since $4 \neq -4$ , the third equation does not hold true, which implies that there is no such scalar $\lambda$ that can make the vectors collinear.

Conclude that there is no value of $k$ that will make the vectors $(2, -6, 4)$ and $(1, k, -2)$ collinear.
The correct answer is: A. There is no such $k$ .