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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Vector Algebra

BITSAT 2023 · Paper 1 · Mathematics · Vector Algebra

If a = i + j + k , a · b =1 and a × b = j - k , then b is

If a⃗=i^+j^+k^,a⃗⋅b⃗=1\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1 and a⃗×b⃗=j⃗−k⃗\vec{a} \times \vec{b}=\vec{j}-\vec{k}, then b⃗\vec{b} is

Options

  1. Ai^−j^+k^\hat{i}-\hat{j}+\hat{k}
  2. B2j^−k^2 \hat{j}-\hat{k}
  3. C2i^2 \hat{i}
  4. Di^\hat{i}

Correct answer

D. i^\hat{i}

Step-by-step solution

Given a⃗=i^+j^+k^,a⃗⋅b⃗=1\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1 and a⃗×b⃗=j⃗−k⃗\vec{a} \times \vec{b}=\vec{j}-\vec{k},
Let bˉ=b1i^+b2j^+b3k^\bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \widehat{k}
Now, \hat{j}-\widehat{k}=\bar{a} \times \bar{b}=\left|i^j^k^111b2b2b3\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 1 & 1 & 1 \\ b_2 & b_2 & b_3\end{array}\right|
⇒b3−b2=0,b1−b3=1,b2−b1=−1\Rightarrow b_3-b_2=0, b_1-b_3=1, b_2-b_1=-1
⇒b3=b2,b1=b2+1\Rightarrow b_3=b_2, b_1=b_2+1
Also, aˉ⋅bˉ=1\bar{a} \cdot \bar{b}=1
⇒b1+b2+b3=1\Rightarrow b_1+b_2+b_3=1
⇒3b2+1=1\Rightarrow 3 b_2+1=1
⇒b2=0\Rightarrow b_2=0
⇒b1=1,b3=0\Rightarrow b_1=1, b_3=0
Thus, bˉ=i^\bar{b}=\hat{i}
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