Chapter 10 · Vector Algebra

Exercise 10.4 · Q5: $\lambda$ ⁠, $\mu$ with $(2\hat i+6\hat j+27\hat k)\times$ $(\hat i+\lambda\hat j+\mu\hat k)=\vec 0$

Exercise 10.4Vector product zero, parallel vectors · Section 10.6.3

The question

Find $\lambda$ and $\mu$ if $(2\hat i + 6\hat j + 27\hat k)\times(\hat i + \lambda\hat j + \mu\hat k) = \vec 0$ ⁠.

Exam answer

Let $\vec a = 2\hat i + 6\hat j + 27\hat k$ and $\vec b = \hat i + \lambda\hat j + \mu\hat k$ ⁠.

\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} \\ &= \hat i(6\mu - 27\lambda) - \hat j(2\mu - 27) \\ &\quad + \hat k(2\lambda - 6) \end{aligned}

Since $\vec a\times\vec b = \vec 0$ ⁠, each component is 0:

\begin{aligned} 2\lambda - 6 = 0 &\Rightarrow \lambda = 3 \\ 2\mu - 27 = 0 &\Rightarrow \mu = \frac{27}{2} \end{aligned}

These also make the $\hat i$ component 0: $6\mu - 27\lambda = 81 - 81 = 0$ ⁠.

$\therefore\ \lambda = 3$ and $\mu = \dfrac{27}{2}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A vector is $\vec 0$ exactly when all three of its components are 0. So write $\vec a\times\vec b$ in components, set each component equal to 0, and solve. Geometrically, a zero vector product of two nonzero vectors means the vectors are parallel.

Given, to find, and the plan

Given: $(2\hat i + 6\hat j + 27\hat k)\times(\hat i + \lambda\hat j + \mu\hat k) = \vec 0$ ⁠.
To find: $\lambda$ and $\mu$ ⁠.
Plan: (1) Expand the vector product as a determinant. (2) Set each component to 0. (3) Solve, and check the third equation. (4) Confirm by seeing the vectors as parallel.

Basics you need

1Vector product in components

$\vec a\times\vec b$ is the determinant with $\hat i, \hat j, \hat k$ on top and the components of $\vec a$ ⁠, then $\vec b$ ⁠, below; expand along the top row with signs $+, -, +$ (Section 10.6.3). Basics: a determinant measures a signed area

2Equal vectors, zero vector

Two vectors are equal exactly when their corresponding components are equal; so $x\hat i + y\hat j + z\hat k = \vec 0$ means $x = y = z = 0$ (Section 10.5.1). Lesson: components

3Zero vector product means parallel

For nonzero $\vec a$ and $\vec b$ ⁠, $\vec a\times\vec b = \vec 0$ if and only if $\vec a$ and $\vec b$ are parallel (Section 10.6.3, Observation 2). Lesson: the cross product

Solution

Expand the vector product
$\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} \\ &= \hat i(6\mu - 27\lambda) - \hat j(2\mu - 27) \\ &\quad + \hat k(2\lambda - 6). \end{aligned}$
Name $\vec a = 2\hat i + 6\hat j + 27\hat k$ and $\vec b = \hat i + \lambda\hat j + \mu\hat k$ ⁠. The minors are $(6)(\mu) - (27)(\lambda)$ ⁠, $(2)(\mu) - (27)(1)$ and $(2)(\lambda) - (6)(1)$ ⁠. The unknowns simply ride along in the determinant like numbers.

The determinant form of $\vec a\times\vec b$ (Section 10.6.3). Basics: a determinant measures a signed area
Every component must be 0
$\begin{aligned} 6\mu - 27\lambda &= 0 &&(\hat i) \\ 2\mu - 27 &= 0 &&(\hat j) \\ 2\lambda - 6 &= 0 &&(\hat k) \end{aligned}$
$\vec a\times\vec b = \vec 0 = 0\hat i + 0\hat j + 0\hat k$ ⁠, and equal vectors have equal components. The $\hat j$ component is $-(2\mu - 27)$ ⁠, and $-(2\mu - 27) = 0$ is the same equation as $2\mu - 27 = 0$ ⁠.

Equality of vectors in components (Section 10.5.1). Lesson: components
Solve, and check the third equation

From the $\hat k$ equation, $\lambda = 3$ ⁠; from the $\hat j$ equation, $\mu = \frac{27}{2}$ ⁠. Then
$\begin{aligned} 6\mu - 27\lambda &= 6 \cdot \frac{27}{2} - 27 \cdot 3 \\ &= 81 - 81 = 0. \end{aligned}$
Three equations in two unknowns can contradict each other; here the third one is satisfied, so $\lambda = 3$ ⁠, $\mu = \frac{27}{2}$ is the one and only solution. (The $\hat j$ and $\hat k$ equations alone already force these values, so there is no other.)

Solving linear equations; checking every equation. Lesson: the cross product, exercises
See it as parallel vectors

With these values, $\vec b = \hat i + 3\hat j + \frac{27}{2}\hat k = \frac12(2\hat i + 6\hat j + 27\hat k) = \frac12\vec a$ ⁠. The ratios of the components agree:
$\frac12 = \frac{\lambda}{6} = \frac{\mu}{27}.$
Two nonzero vectors have a zero vector product exactly when they are parallel, that is, when one is a scalar multiple of the other. Comparing the $\hat i$ components, the multiple must be $\frac12$ ⁠, and then $\lambda = \frac12 \cdot 6 = 3$ and $\mu = \frac12 \cdot 27 = \frac{27}{2}$ ⁠: the same answer by a second route.

$\vec a\times\vec b = \vec 0$ for nonzero vectors iff they are parallel (Section 10.6.3, Observation 2). Lesson: the cross product

Result

$\lambda = 3$ and $\mu = \frac{27}{2}$ ⁠.

(Textbook answer: $3, \frac{27}{2}$ ⁠.)

Check it with numbers

$\mu = \frac{27}{2} = 13.5$ ⁠. With $\vec b = \hat i + 3\hat j + 13.5\hat k$ ⁠:

$\hat i$ ⁠: $(6)(13.5) - (27)(3) = 81 - 81 = 0$ ⁠; $\hat j$ ⁠: $(2)(13.5) - (27)(1) = 27 - 27 = 0$ ⁠; $\hat k$ ⁠: $(2)(3) - (6)(1) = 6 - 6 = 0$ ⁠. So $\vec a\times\vec b = \vec 0$ ⁠.

Common mistakes

Solving only two equations. The $\hat i$ equation $6\mu = 27\lambda$ must also hold; checking it confirms the answer.
Turning the ratios upside down. $\vec b = \frac12\vec a$ ⁠, so $\lambda = \frac12 \cdot 6$ ⁠; writing $\frac{\lambda}{6} = \frac21$ gives $\lambda = 12$ ⁠, which fails the $\hat k$ equation.
Worrying about the minus on the $\hat j$ term. $-(2\mu - 27) = 0$ and $2\mu - 27 = 0$ are the same equation.

What to take away

$\vec a\times\vec b = \vec 0$ gives three equations, one for each component.
For nonzero vectors it means $\vec a \parallel \vec b$ ⁠: proportional components.
Check the equation you did not use to solve.

Every fact used, and where it comes from

Fact	Used in	Source
$\vec a\times\vec b$ as a determinant in components	Step 1	Section 10.6.3 (Basics: a determinant measures a signed area)
Vectors are equal iff their components are equal	Step 2	Section 10.5.1 (Lesson: components)
$\vec a\times\vec b = \vec 0$ for nonzero vectors iff they are parallel	Step 4	Section 10.6.3, Observation 2 (Lesson: the cross product)
Parallel vectors: $\vec b = k\vec a$ for a scalar $k$	Step 4	Section 10.5
The question and the answer	Result	Exercise 10.4 Q5; NCERT answer key (Lesson: the cross product, exercises)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 10.4 · Q5: λ\lambdaλ⁠, μ\muμ with (2i^+6j^+27k^)×(2\hat i+6\hat j+27\hat k)\times(2i^+6j^​+27k^)× (i^+λj^+μk^)=0⃗(\hat i+\lambda\hat j+\mu\hat k)=\vec 0(i^+λj^​+μk^)=0

Basics you need

Solution

Expand the vector product

Every component must be 0

Solve, and check the third equation

See it as parallel vectors

Every fact used, and where it comes from

Exercise 10.4 · Q5: $\lambda$ ⁠, $\mu$ with $(2\hat i+6\hat j+27\hat k)\times$ $(\hat i+\lambda\hat j+\mu\hat k)=\vec 0$