Three Dimensional Geometry — JEE MCQs with answers

Answer: B — If $a, b, c$ are direction ratios of a line, then $ka, kb, kc$ with $k \ne 0$ is also a set of direction ratios, and any two sets are proportional. Option D is $3, -6, 9$ multiplied by $\dfrac{1}{3}$, option C by $-2$ and option A by $\dfrac{1}{6}$, so all three are sets of direction ratios of the line. In option B the first two entries are $3, -6$ multiplied by $-1$ but the last entry is $9$ multiplied by $1$; no single multiplier works, so $-3, 6, 9$ is not a set of direction ratios of this line.

Answer: B — By the Remark, direction ratios $2, -3, 6$ mean the line is parallel to $\vec{b} = 2\hat{i} - 3\hat{j} + 6\hat{k}$. The point $(-1, 0, 4)$ has position vector $\vec{a} = -\hat{i} + 0\hat{j} + 4\hat{k} = -\hat{i} + 4\hat{k}$. So $\vec{r} = \vec{a} + \lambda\vec{b}$ is option B. Option D drops the zero and attaches $4$ to $\hat{j}$, which is the point $(-1, 4, 0)$. Option C interchanges the point and the direction. Option A makes both slips.

Answer: D — Here $(x_1, y_1, z_1) = (-3,\ -3,\ -4)$ and $(x_2, y_2, z_2) = (1,\ 4,\ -2)$; the direction ratios are $2,\ 1,\ -2$ for $l_1$ and $4,\ 4,\ 1$ for $l_2$. The first row of the determinant is $4,\ 7,\ 2$. Expanding along the first row, the determinant is $(4)(9) - (7)(10) + (2)(4) = -26$. This is option D. Option A takes the first row as $x_1 - x_2,\ y_1 - y_2,\ z_1 - z_2$, that is $-4,\ -7,\ -2$, which reverses the sign of the determinant. Option B adds the middle term of the expansion instead of subtracting it: $(4)(9) + (7)(10) + (2)(4) = 114$. Option C makes both slips: the first row $-4,\ -7,\ -2$ and the middle term added.

Answer: C — The direction vectors are the ones multiplied by the parameters: $\vec{b}_1 = -2\hat{i} + 4\hat{j} - 4\hat{k}$ and $\vec{b}_2 = 2\hat{i} - 2\hat{k}$. $\vec{b}_1\cdot\vec{b}_2 = (-2)(2) + (4)(0) + (-4)(-2) = 4$, $|\vec{b}_1| = \sqrt{4 + 16 + 16} = 6$ and $|\vec{b}_2| = \sqrt{4 + 0 + 4} = 2 \sqrt{2}$. The correct value is $\cos\theta = \dfrac{|4|}{6 \cdot 2 \sqrt{2}} = \dfrac{\sqrt{2}}{6}$. Each slip in turn: with the position vectors $2\hat{j} - 5\hat{k}$ and $4\hat{i} + 2\hat{j} + \hat{k}$ (scalar product $-1$, magnitudes $\sqrt{29}$ and $\sqrt{21}$) the value is $\dfrac{\sqrt{609}}{609}$ (option B); dividing by $6 \cdot 8$ gives $\dfrac{1}{12}$ (option A); with the $\hat{k}$-component of $\vec{b}_2$ taken as $2$ the scalar product is $-12$ and the value is $\dfrac{\sqrt{2}}{2}$ (option C); dividing by $2 \sqrt{2}$ alone gives $\sqrt{2}$ (option D). Only option C gives $\dfrac{\sqrt{2}}{2}$.

Answer: B — On p.377 the opening line is printed with the attribution A. DeMorgan (in capitals, as A.DEMORGAN). Leonhard Euler is the mathematician whose portrait appears on the same page, not the person the line is attributed to; the other two names do not appear on the page.

Answer: D — Right after obtaining the three equations the textbook says 'These are parametric equations of the line'; $\lambda$ is the parameter. That is option D. 'The Cartesian equation of the line' (option C) is the name it gives to the next result, $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$, which comes only after $\lambda$ is eliminated. The vector equation (option A) is $\vec{r} = \vec{a} + \lambda\vec{b}$, the equation the substitution starts from. The direction ratios (option B) are the three numbers $a, b, c$, not the equations.

Answer: C — The three angles are equal, so $l = m = n$, and $l^2 + m^2 + n^2 = 1$ gives $3l^2 = 1$, $l = \pm\dfrac{1}{\sqrt{3}}$. An obtuse angle has a negative cosine, so $l = m = n = -\dfrac{1}{\sqrt{3}}$: option C. Option A takes the positive root; those are the direction cosines of the same line directed the opposite way, which makes equal acute angles with the axes. Option B uses $l + m + n = 1$ in place of the sum of squares and then attaches the negative sign; option D is the same slip without the sign. In options B and D the squares add up to $\dfrac{1}{3}$, not 1.

Answer: B — Direction ratios of AB are $3 - 1,\ 5 - 2,\ 2 - 3$, i.e. $2,\ 3,\ -1$ (Note on page 380). For $D(x, y, z)$ the line CD has direction ratios $x - 2,\ y + 1,\ z - 4$, and the lines are perpendicular when $2(x - 2) + 3(y + 1) - (z - 4) = 0$ (page 384). $D(3, 0, 9)$: CD has direction ratios $1, 1, 5$ and $2 + 3 - 5 = 0$, so this point works. $D(3, 0, -1)$: direction ratios $1, 1, -5$ give $2 + 3 + 5 = 10$; it works only if the third direction ratio of AB is taken as $+1$. $D(1, 1, 5)$: direction ratios $-1, 2, 1$ give $-2 + 6 - 1 = 3$; here it is the coordinates of D themselves, not the differences, that satisfy $2x + 3y - z = 0$. $D(1, 1, -5)$: direction ratios $-1, 2, -9$ give $-2 + 6 + 9 = 13$.

Answer: B — The text takes 'any point S on $l_1$ with position vector $\vec{a}_1$ and T on $l_2$ with position vector $\vec{a}_2$'. In the figure the shortest distance segment runs from P on $l_1$ (where the right angle is marked) up to Q on $l_2$. So S and P are on $l_1$ and T and Q on $l_2$, option B. Option A interchanges P and Q, although the right angle with $l_1$ is at P. Option C interchanges S and T, although $\vec{a}_1$ belongs to $l_1$. Option D interchanges both pairs.

Answer: D — The footnote on p.377 reads that for various activities in three dimensional geometry one may refer to the book "A Hand Book for designing Mathematics Laboratory in Schools", NCERT, 2005. It is attached to the sentence about the vector approach making the study simple and elegant. The other three options are topics named in the body of the Introduction, not what the footnote points to.