Given that a,b\mathbf{a}, \mathbf{b} and c\mathbf{c} are non-zero vectors such that (a+b)×(a+c)=b×c(\mathbf{a}+\mathbf{b})\times(\mathbf{a}+\mathbf{c}) = \mathbf{b} \times \mathbf{c}, and b≠c\mathbf{b} \ne \mathbf{c}, find the relationship between a,b\mathbf{a}, \mathbf{b} and c\mathbf{c}. It is given instead that a,b\mathbf{a}, \mathbf{b} and c\mathbf{c} satisfy the equation a+b+c=0\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0} with ∣a∣=2|\mathbf{a}| = 2, ∣b∣=3|\mathbf{b}| = 3 and ∣c∣=4|\mathbf{c}| = 4. Find the value of a⋅b+b⋅c+c⋅a\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}.

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

4(a)

1. Expand the cross product on the left hand side of the equation.

(a+b)×(a+c)=a×a+a×c+b×a+b×c(\mathbf{a}+\mathbf{b})\times(\mathbf{a}+\mathbf{c}) = \mathbf{a}\times\mathbf{a} + \mathbf{a}\times\mathbf{c} + \mathbf{b}\times\mathbf{a} + \mathbf{b}\times\mathbf{c}

2. Simplify the equation using the property $\mathbf{a} \times \mathbf{a} = \mathbf{0}$.

0+a×c+b×a+b×c=b×c\mathbf{0} + \mathbf{a}\times\mathbf{c} + \mathbf{b}\times\mathbf{a} + \mathbf{b}\times\mathbf{c} = \mathbf{b}\times\mathbf{c}
a×c+b×a=0\mathbf{a}\times\mathbf{c} + \mathbf{b}\times\mathbf{a} = \mathbf{0}

3. Rearrange and factorise the expression using anti-commutativity of the cross product.

Since b×a=−a×b\mathbf{b}\times\mathbf{a} = -\mathbf{a}\times\mathbf{b}:
a×c−a×b=0\mathbf{a}\times\mathbf{c} - \mathbf{a}\times\mathbf{b} = \mathbf{0}
a×(c−b)=0\mathbf{a}\times(\mathbf{c} - \mathbf{b}) = \mathbf{0}

4. Deduce the relationship between the vectors.

We are given that a\mathbf{a} is a non-zero vector, and b≠c\mathbf{b} \ne \mathbf{c} implies c−b≠0\mathbf{c} - \mathbf{b} \ne \mathbf{0}.
Since the cross product of two non-zero vectors is the zero vector, they must be parallel.
Therefore, a\mathbf{a} is parallel to c−b\mathbf{c} - \mathbf{b}.

Answer

a\mathbf{a} is parallel to c−b\mathbf{c} - \mathbf{b}

4(b)

1. Take the dot product of the given vector sum with itself.

(a+b+c)⋅(a+b+c)=0⋅0=0(\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) = \mathbf{0} \cdot \mathbf{0} = 0

2. Expand the dot product using the distributive property and the definition of magnitude.

∣a∣2+∣b∣2+∣c∣2+2(a⋅b+b⋅c+c⋅a)=0|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 0

3. Substitute the given magnitudes to find the value of the required expression.

(2)2+(3)2+(4)2+2(a⋅b+b⋅c+c⋅a)=0(2)^2 + (3)^2 + (4)^2 + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 0
4+9+16+2(a⋅b+b⋅c+c⋅a)=04 + 9 + 16 + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 0
29+2(a⋅b+b⋅c+c⋅a)=029 + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 0
a⋅b+b⋅c+c⋅a=−292=−14.5\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} = -\frac{29}{2} = -14.5

Answer

−14.5-14.5

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1The lines l1l_1 and l2l_2 have equations r=(4110)+λ(1−1−6),λ∈R\mathbf{r} = \begin{pmatrix} 4 \\ 1 \\ 10 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix}, \lambda \in \mathbb{R}…
  2. 2The vectors a\mathbf{a} and b\mathbf{b} are given by a=i+(sin⁡θ)j+(cos⁡θ)k\mathbf{a}=\mathbf{i}+(\sin\theta)\mathbf{j}+(\cos\theta)\mathbf{k}\quad…
  3. 3With reference to the origin OO, the points AA, PP and QQ have position vectors a\mathbf{a}, p\mathbf{p} and q\mathbf{q}…
  4. 4A plane Π1\Pi_1 has equation x+y+z=3x + y + z = 3. A line passes through the points PP and QQ with position vectors i+j+2k\mathbf{i} + \mathbf{j} + 2\mathbf{k}…
  5. 5The planes Π1\Pi_1 and Π2\Pi_2 have vector equations r=λ1(i+j−k)+μ1(2i−j+k)andr=λ2(i+2j+k)+μ2(3i+j−k)\mathbf{r} = \lambda_1(\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mu_1(2\mathbf{i} - \mathbf{j} + \mathbf{k}) \quad \text{and} \quad \mathbf{r} = \lambda_2(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \mu_2(3\mathbf{i} + \mathbf{j} - \mathbf{k})…