Let a⃗=-î-k̂,b⃗=-î+ĵ and c⃗=î+2ĵ+3k̂ be three given vectors. If r⃗ is a vector such that r⃗×b⃗=c⃗×b⃗ and r⃗·a⃗=0, then the value of r⃗·b⃗ is

Let a⃗=-î-k̂,b⃗=-î+ĵ and c⃗=î+2ĵ+3k̂ be three given vectors. If r⃗ is a vector such that r⃗×b⃗=c⃗×b⃗ and r⃗·a⃗=0, then the value of r⃗·b⃗ is Given: a⃗=-î-k̂ b⃗=-î+ĵ c⃗=î+2ĵ+3k̂ Condition: r⃗×b⃗=c⃗×b⃗ r⃗·a⃗=0 To find: r⃗·b⃗ c⃗ = î + 2ĵ + 3k̂ = (1, 2, 3) r⃗×b⃗ = c⃗×b⃗ ⇒ (r⃗-c⃗)×b⃗ = 0⃗ (r⃗-c⃗) ∥ b⃗ ⇒ r⃗ = c⃗ + λb⃗ Plan: 1. r⃗ = c⃗ + λb⃗ 2. (c⃗ + λb⃗)·a⃗ = 0 ⇒ λ 3. r⃗·b⃗ a⃗·b⃗ = (-1)(-1) + (0)(1) + (-1)(0) a⃗·b⃗ = 1 c⃗·a⃗ = (1)(-1) + (2)(0) + (3)(-1) c⃗·a⃗ = -4 r⃗·a⃗ = 0 ⇒ (c⃗ + λb⃗)·a⃗ = 0 c⃗·a⃗ + λ(b⃗·a⃗) = 0 (c⃗·a⃗) + λ(b⃗·a⃗) = 0 -4 + λ(1) = 0 ⇒ λ = 4 r⃗ = c⃗ + 4b⃗ = (1, 2, 3) + 4(-1, 1, 0) r⃗ = (-3, 6, 3) r⃗·a⃗ = (-3)(-1) + (6)(0) + (3)(-1) = 3 - 3 = 0 checkmark r⃗·b⃗ = (-3)(-1) + (6)(1) + (3)(0) r⃗·b⃗ = 3 + 6 = 9 c⃗·b⃗ = (1)(-1) + (2)(1) + (3)(0) = 1 |b⃗|² = (-1)² + 1² + 0² = 2 r⃗·b⃗ = 1 + 4(2) = 9 checkmark r⃗·b⃗ = 9

Let $\vec{a}=-\hat{i}-\hat{k},\vec{b}=-\hat{i}+\hat{j}$ and $\vec{c}=\hat{i}+2\hat{j}+3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}\times\vec{b}=\vec{c}\times\vec{b}$ and $\vec{r}\cdot\vec{a}=0$ , then the value of $\vec{r}\cdot\vec{b}$ is

Answer: $9$

Vector Algebra Let

\vec{a}=-\hat{i}-\hat{k},\vec{b}=-\hat{i}+\hat{j}

and

\vec{c}=\hat{i}+2\hat{j}+3\hat{k}

be three given vectors. If

\vec{r}

is a vector such that

\vec{r}\times\vec{b}=\vec{c}\times\vec{b}

and

\vec{r}\cdot\vec{a}=0

, then the value of

\vec{r}\cdot\vec{b}

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Written solution

$\text{Given: } \vec{a}=-\hat{i}-\hat{k}$ $\vec{b}=-\hat{i}+\hat{j}$ $\vec{c}=\hat{i}+2\hat{j}+3\hat{k}$ $\text{Condition: } \vec{r}\times\vec{b}=\vec{c}\times\vec{b}$ $\vec{r}\cdot\vec{a}=0$ $\text{To find: } \vec{r}\cdot\vec{b}$ $\vec{c} = \hat{i} + 2\hat{j} + 3\hat{k} = (1,\ 2,\ 3)$ $\vec{r}\times\vec{b} = \vec{c}\times\vec{b}$ $\Rightarrow (\vec{r}-\vec{c})\times\vec{b} = \vec{0}$ $(\vec{r}-\vec{c}) \parallel \vec{b} \ \Rightarrow \ \vec{r} = \vec{c} + \lambda\vec{b}$ $\text{Plan: } 1.\ \vec{r} = \vec{c} + \lambda\vec{b}$ $2.\ (\vec{c} + \lambda\vec{b})\cdot\vec{a} = 0 \Rightarrow \lambda$ $3.\ \vec{r}\cdot\vec{b}$ $\vec{a}\cdot\vec{b} = (-1)(-1) + (0)(1) + (-1)(0)$ $\vec{a}\cdot\vec{b} = 1$ $\vec{c}\cdot\vec{a} = (1)(-1) + (2)(0) + (3)(-1)$ $\vec{c}\cdot\vec{a} = -4$ $\vec{r}\cdot\vec{a} = 0 \Rightarrow (\vec{c} + \lambda\vec{b})\cdot\vec{a} = 0$ $\vec{c}\cdot\vec{a} + \lambda(\vec{b}\cdot\vec{a}) = 0$ $(\vec{c}\cdot\vec{a}) + \lambda(\vec{b}\cdot\vec{a}) = 0$ $-4 + \lambda(1) = 0 \Rightarrow \lambda = 4$ $\vec{r} = \vec{c} + 4\vec{b} = (1,\ 2,\ 3) + 4(-1,\ 1,\ 0)$ $\vec{r} = (-3,\ 6,\ 3)$ $\vec{r}\cdot\vec{a} = (-3)(-1) + (6)(0) + (3)(-1) = 3 - 3 = 0 \ \checkmark$ $\vec{r}\cdot\vec{b} = (-3)(-1) + (6)(1) + (3)(0)$ $\vec{r}\cdot\vec{b} = 3 + 6 = 9$ $\vec{c}\cdot\vec{b} = (1)(-1) + (2)(1) + (3)(0) = 1$ $|\vec{b}|^2 = (-1)^2 + 1^2 + 0^2 = 2$ $\vec{r}\cdot\vec{b} = 1 + 4(2) = 9 \ \checkmark$ $\vec{r}\cdot\vec{b} = 9$

Written solution

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