Applications of Vector Algebra - Book-Back Answer Key
Applications of Vector Algebra — Book-Back Answer Key (Exercise 6.10)
Correct option and a one-line reason for each of the 25 multiple-choice questions. Every answer was independently verified against the worked solution.
| Q.No | Answer | Reason |
|---|---|---|
| 1 | D | parallel vectors give \(\vec{a}\times\vec{b}=\vec{0}\), so the box product is \(0\). |
| 2 | C | \(\vec{\alpha}\) coplanar with \(\vec{\beta},\vec{\gamma}\) makes \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\). |
| 3 | A | mutually perpendicular vectors give \(|[\vec{a},\vec{b},\vec{c}]|=|\vec{a}||\vec{b}||\vec{c}|\). |
| 4 | B | \((\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}=1\cdot\vec{b}-0=\vec{b}\). |
| 5 | A | two cyclic terms give \(+1\), the odd permutation gives \(-1\): \(1+1-1=1\). |
| 6 | C | the determinant of the three edges equals \(\pi\). |
| 7 | A | \(\sin^{2}\theta=\tfrac14\Rightarrow\theta=\tfrac{\pi}{6}\). |
| 8 | A | \(\lambda=-1,\ \mu=1\Rightarrow\lambda+\mu=0\). |
| 9 | A | \([\vec{a}\times\vec{b},\vec{b}\times\vec{c},\vec{c}\times\vec{a}]=3^{2}=9\), squared \(=81\). |
| 10 | B | \(\cos\theta=-\tfrac{1}{\sqrt2}\Rightarrow\theta=\tfrac{3\pi}{4}\). |
| 11 | C | volume \(=(V^{2})^{2}=8^{2}=64\) cubic units. |
| 12 | A | the two plane normals are parallel, so the angle is \(0\). |
| 13 | B | the identity collapses to \(\vec{c}\parallel\vec{a}\). |
| 14 | D | \(22\vec{b}-13\vec{c}=-17\hat{i}-21\hat{j}-97\hat{k}\). |
| 15 | D | direction ratios \((3,-2,0)\) and \((2,3,4)\) are perpendicular. |
| 16 | B | \(\alpha=-6\) from perpendicularity, \(\beta=7\) from the point \((2,1,-2)\). |
| 17 | C | \(\sin\theta=\tfrac{1}{\sqrt2}\Rightarrow\theta=45^{\circ}\). |
| 18 | D | \(t=1\) gives the point \((5,-1,1)\). |
| 19 | B | \(\tfrac{7}{\sqrt{49}}=1\). |
| 20 | A | \(\tfrac{7}{2\sqrt{14}}=\tfrac{\sqrt7}{2\sqrt2}\). |
| 21 | B | \(\tfrac{3}{c^{2}}=1\Rightarrow c=\pm\sqrt3\). |
| 22 | C | \(t=0\) and \(t=1\) give \((1,-2,-1)\) and \((1,4,-2)\). |
| 23 | D | \(|3+k|=6\Rightarrow k=3\) or \(-9\). |
| 24 | C | proportional normals give \(\lambda=-\tfrac12,\ \mu=-2\). |
| 25 | A | \(13+\lambda^{2}=25\Rightarrow\lambda=2\sqrt3\). |