Let a = i + j + k, b = 2i + 4j - 5k and c = xi + 2j + 3k, x ∈ R. If d is the unit vector in the direction of

Question

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{{c}}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}\). If \(\vec{d}\) is the unit vector in the direction of \(\vec{b}+\vec{c}\) such that \(\vec{a} \cdot \vec{d}=1\), then \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is equal to

(1) 9

(2) 6

(3) 3

(4) 11

Answer 1

Correct option is (4) 11

\(\vec{\mathrm{d}}=\lambda(\vec{\mathrm{b}}+\vec{\mathrm{c}})\)

\(\vec{a} . \vec{d}=\lambda(\vec{b} . \vec{a}+\vec{c} . \vec{a})\)

\(1=\lambda(1+\mathrm{x}+5)\)

\(1=\lambda(\mathrm{x}+6)\)

\(|\vec{\mathrm{d}}|=1 \quad \frac{1}{\lambda}=\mathrm{x}+6\)

\(|\lambda(\vec{\mathrm{b}}+\vec{\mathrm{c}})|=1\)

\(|\lambda((x+2) \hat{i}+6 \hat{j}-2 \hat{k})|=1\)

\(\lambda^{2}\left((x+2)^{2}+6^{2}+2^{2}\right)=1\)

\(\mathrm{x}^{2}+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^{2}\)

\(\mathrm{x}^{2}+4 \mathrm{x}+44=\mathrm{x}^{2}+12 \mathrm{x}+36\)

\(8 \mathrm{x}=8, \mathrm{x}=1\)

\(\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 4 & -5 \\ x & 2 & 3\end{array}\right|=(\vec{a} \times \vec{b}) \cdot \vec{c}\)

\(\left|\begin{array}{ccc}0 & 0 & 1 \\ -2 & 9 & -4 \\ x-2 & -1 & 3\end{array}\right|=2-9(x-2)\)

\(=20-9 \mathrm{x}\)

At \(\mathrm{x}=1\)

\(20-9=11\)