Let vector a=i+j+k, vector b=2i+2j+k and vector d=vector a x vector b.

Question

Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}.\) If \(\vec{c}\) is a vector such that \(\vec{a} \cdot \vec{c}=|\vec{c}|,|\overrightarrow{\mathrm{c}}-2 \vec{a}|^{2}=8\) and the angle between \(\overrightarrow{\mathrm{d}}\) and \(\overrightarrow{\mathrm{c}}\) is \(\frac{\pi}{4},\) then \(|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^{2}\) is equal to \(\ldots .\)

Answer 1

Answer is: 6

\(\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}\)

\(\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)

\(\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\)

\(=-\hat{i}+\hat{j}\)

\(|\overrightarrow{\mathbf{c}}-2 \overrightarrow{\mathrm{a}}|^{2}=8\)

\(|c|^{2}+4|a|^{2}-4(a . c)=8\)

\(c^{2}+12-4 c=8\)

\(c^{2}-4 \mathrm{c}+4=0\)

\(|c|=2\)

\(\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\)

\(\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}\)

\(\left(|\mathrm{d}||\mathrm{c}| \sin \frac{\pi}{4}\right)^{2}=((\mathrm{a} . \mathrm{c}) \cdot \mathrm{b}-(\mathrm{b} . \mathrm{c}) \cdot \mathrm{a})^{2}\)

\(4=4 b^{2}+(b . c)^{2} 2(a)^{2}-2(b . c)(a . b)  \)   

Let b. c = x

\(4=36+3 x^{2}-20 x\)

\(3 x^{2}-20 x+32=0\)

\(3 x^{2}-12 x-8 x+32=0\)

\(\mathrm{x}=\frac{8}{3}, 4\)

b.c \(=\frac{8}{3}, 4\)

b.c \(=\frac{8}{3}\)

Now \(|10-3 \mathrm{~b} . \mathrm{c}|+|\mathrm{d} \times \mathrm{c}|^{2}\)

\(|10-8|+(2)^{2}\)

\(\Rightarrow 6\)