Exam Questions & Step-by-Step Markschemes (5 Problems)
Question 1 • 2024
3 Marks
If four points with position vectors \vec{a} = 6\hat{i} + 3\hat{j} - 4\hat{k} , \vec{b} = \hat{i} - 2\hat{j} + 3\hat{k} , \vec{c} = 3\hat{i} + \lambda \hat{j} + 4\hat{k} , and \vec{d} = -4\hat{i} + 4\hat{j} - 4\hat{k} are coplanar, find the value of \lambda .
Answer: \lambda = -\frac{209}{70}
Question 2 • 2023
4 Marks
Find the shortest distance between the lines: \vec{r}_1 = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k}) \vec{r}_2 = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})
Answer: d = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19} \text{ units}
Question 3 • 2024
4 Marks
Show that the lines \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 4}{5} = \frac{y - 1}{2} = z are coplanar. Also, find the equation of the plane containing them.
Answer: 5x - 18y + 11z - 2 = 0 \quad (\text{Lines are coplanar})
Question 4 • 2024
3 Marks
Find the area of a parallelogram whose diagonals are determined by the vectors \vec{d}_1 = 3\hat{i} + \hat{j} - 2\hat{k} and \vec{d}_2 = \hat{i} - 3\hat{j} + 4\hat{k} .
Answer: \text{Area} = 5\sqrt{3} \text{ square units}
Question 5 • 2024
3 Marks
સદિશ \vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} નો સદિશ \vec{b} = \hat{i} + 2\hat{j} + \hat{k} પરનો પ્રક્ષેપ (projection) શોધો.
Answer: \text{Projection} = \frac{5\sqrt{6}}{3}