Let {a} be a non-zero vector parallel to the line of intersection of the two places described by {i}+{j}, {i}+{k} and {i}-{j}, {j}-{k}. If θ is the angle between the vector {a} and the vector {b}=2{i}-2{j}+{k} and {a}· {b}=6, then ordered pair (θ, |{a}× {b}|) is equal to : [11-Apr-2023 shift 1]

Correct Answer is : (π4,6)≤ft(π/4, 6)(4π,6)

Correct Answer is : (π4,6)≤ft(π/4, 6)(4π,6)

Test Index

Vector Algebra Part 2

Section: Mathematics

Question : 15 of 100

Marks: +1, -0

Let

\vec{a}

\hat{i}+\hat{j}, \hat{i}+\hat{k}

\hat{i}-\hat{j}, \hat{j}-\hat{k}

\theta

\vec{a}

\vec{b}=2\hat{i}-2\hat{j}+\hat{k}

\vec{a}\cdot \vec{b}=6

(\theta, |\vec{a}\times \vec{b}|)

Solution:

\vec{n}_1

and

\vec{n}_2

are normal vector to the plane

\hat{i}+\hat{j}, \hat{i}+\hat{k}

and

\hat{i}-\hat{j}, \hat{i}-\hat{k}

respectively

\vec{n}_1=\begin{vmatrix} \hat{i} & \hat{j} & \hat{j} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix}=\hat{i}-\hat{j}-\hat{k}

\vec{n}_2=\begin{vmatrix} \hat{i} & \hat{j} & \hat{j} \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{vmatrix}=\hat{i}+\hat{j}+\hat{k}

\vec{a}=\lambda\left|\vec{n}_2\times \vec{n}_2\right|

=\lambda\begin{vmatrix} i & \hat{j} & \hat{j} \\ 1 & -1 & -1 \\ 1 & 1 & -1 \end{vmatrix}=\lambda(-2\hat{j}+2\hat{k})

\;\vec{a}\cdot \vec{b}=\lambda\left|0+4+2\right|=6

\;\Rightarrow \lambda=1

\;\vec{\alpha}=-2\hat{j}+2\hat{k}

\;\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|a||b|}

\;\cos\theta=\frac{6}{2\sqrt{2}\times 3}=\frac{1}{\sqrt{2}}

\;\theta=\frac{\pi}{4}

\;\text{Now}\;\left|\vec{a}\cdot\vec{b}\right|^2+\left|\vec{a}\times\vec{b}\right|^2=|a|^2|b|^2

\;36+\left|\vec{a}\times b^2\right|=8\times 9=72

\;\left|\vec{a}\times b\right|^2=36

\;\left|\vec{a}\times\vec{b}\right|=6

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