10 Most SAQ’s of Addition of Vectors Chapter in Inter 1st Year Maths-1A (TS/AP)

4 Marks

SAQ-1 : Show that the four points -¯a + 4¯b – ¯3c, ¯3a + 2¯b – 5¯c, -3¯a + 8¯b – ¯5c, -3¯a + 2¯b + ¯c are coplanar, where ¯a, ¯b, ¯c are non-coplanar vectors

$$\overline{OP} = -\overline{a} + 4\overline{b} – 3\overline{c}$$

$$\overline{OQ} = 3\overline{a} + 2\overline{b} – 5\overline{c}$$

$$\overline{OR} = -3\overline{a} + 8\overline{b} – 5\overline{c}$$

$$\overline{OS} = -3\overline{a} + 2\overline{b} + \overline{c}$$

$$\text{where } O \text{ is the origin}$$

$$\text{Calculate } \overline{PQ}:$$

$$\overline{PQ} = \overline{OQ} – \overline{OP}$$

$$= (3\overline{a} + 2\overline{b} – 5\overline{c}) – (-\overline{a} + 4\overline{b} – 3\overline{c})$$

$$= 4\overline{a} – 2\overline{b} – 2\overline{c}$$

$$\text{Calculate } \overline{PR}:$$

$$\overline{PR} = \overline{OR} – \overline{OP}$$

$$= (-3\overline{a} + 8\overline{b} – 5\overline{c}) – (-\overline{a} + 4\overline{b} – 3\overline{c})$$

$$= -2\overline{a} + 4\overline{b} – 2\overline{c}$$

$$\text{Calculate } \overline{PS}:$$

$$\overline{PS} = \overline{OS} – \overline{OP}$$

$$= (-3\overline{a} + 2\overline{b} + \overline{c}) – (-\overline{a} + 4\overline{b} – 3\overline{c})$$

$$= -2\overline{a} – 2\overline{b} + 4\overline{c}$$

$$\text{Determine if } \overline{PQ}, \overline{PR}, \overline{PS} \text{ are coplanar by calculating the determinant:}$$

$$[\overline{PQ} \ \overline{PR} \ \overline{PS}] = \left| \begin{matrix} 4 & -2 & -2 \ -2 & 4 & -2 \ -2 & -2 & 4 \end{matrix} \right| [\overline{a} \ \overline{b} \ \overline{c}]$$

$$= \left[ 4(16 – 4) + 2(-8 – 4) – 2(4 + 8) \right] [\overline{a} \ \overline{b} \ \overline{c}]$$

$$= \left[ 4(12) + 2(-12) – 2(12) \right] [\overline{a} \ \overline{b} \ \overline{c}]$$

$$= \left[ 48 – 24 – 24 \right] [\overline{a} \ \overline{b} \ \overline{c}] = 0 \times [\overline{a} \ \overline{b} \ \overline{c}] = 0$$

$$\text{So, } \overline{PQ}, \overline{PR}, \overline{PS} \text{ are coplanar}$$

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SAQ-2 : Show that the four points 6¯a + 2¯b – ¯c, 2¯a – ¯b + 3¯c, -¯a + 2¯b – 4¯c, – 12¯a – ¯b – 3¯c are coplanar, where ¯a, ¯b, ¯c are non-coplanar vectors

$$\overline{OP} = 6\overline{a} + 2\overline{b} – \overline{c}$$

$$\overline{OQ} = 2\overline{a} – \overline{b} + 3\overline{c}$$

$$\overline{OR} = -\overline{a} + 2\overline{b} – 4\overline{c}$$

$$\overline{OS} = -12\overline{a} – \overline{b} – 3\overline{c}$$

$$\text{where } O \text{ is the origin}$$

$$\overline{PQ} = \overline{OQ} – \overline{OP}$$

$$= (2\overline{a} – \overline{b} + 3\overline{c}) – (6\overline{a} + 2\overline{b} – \overline{c})$$

$$= -4\overline{a} – 3\overline{b} + 4\overline{c}$$

$$\overline{PR} = \overline{OR} – \overline{OP}$$

$$= (-\overline{a} + 2\overline{b} – 4\overline{c}) – (6\overline{a} + 2\overline{b} – \overline{c})$$

$$= -7\overline{a} – 3\overline{c}$$

$$\overline{PS} = \overline{OS} – \overline{OP}$$

$$= (-12\overline{a} – \overline{b} – 3\overline{c}) – (6\overline{a} + 2\overline{b} – \overline{c})$$

$$= -18\overline{a} – 3\overline{b} – 2\overline{c}$$

$$\text{Now, } [\overline{PQ} \ \overline{PR} \ \overline{PS}] = \left| \begin{matrix} -4 & -3 & 4 \ -7 & 0 & -3 \ -18 & -3 & -2 \end{matrix} \right| [\overline{a} \ \overline{b} \ \overline{c}]$$

$$= [-4(0 – 9) + 3(14 – 54) + 4(21 – 0)] [\overline{a} \ \overline{b} \ \overline{c}]$$

$$= [-4(-9) + 3(-40) + 4(21)] [\overline{a} \ \overline{b} \ \overline{c}]$$

$$= [36 – 120 + 84] [\overline{a} \ \overline{b} \ \overline{c}]$$

$$= 0 \times [\overline{a} \ \overline{b} \ \overline{c}] = 0$$

$$\text{So, } \overline{PQ}, \overline{PR}, \overline{PS} \text{ are coplanar}$$

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SAQ-3 : If ¯i, ¯j, ¯k are unit vectors along the positive directions of the coordinate axes then show that the four points 4¯i + 5¯j + ¯k, -¯j – ¯k, 3¯i + 9¯j + 4¯k, -4¯i + 4¯j + 4¯k are coplanar

$$\overline{OP} = 4\overline{i} + 5\overline{j} + \overline{k}$$

$$\overline{OQ} = -\overline{j} – \overline{k}$$

$$\overline{OR} = 3\overline{i} + 9\overline{j} + 4\overline{k}$$

$$\overline{OS} = -4\overline{i} + 4\overline{j} + 4\overline{k}$$

$$\text{where } O \text{ is the origin}$$

$$\overline{PQ} = \overline{OQ} – \overline{OP}$$

$$= (-\overline{j} – \overline{k}) – (4\overline{i} + 5\overline{j} + \overline{k})$$

$$= -4\overline{i} – 6\overline{j} – 2\overline{k}$$

$$\overline{PR} = \overline{OR} – \overline{OP}$$

$$= (3\overline{i} + 9\overline{j} + 4\overline{k}) – (4\overline{i} + 5\overline{j} + \overline{k})$$

$$= -\overline{i} + 4\overline{j} + 3\overline{k}$$

$$\overline{PS} = \overline{OS} – \overline{OP}$$

$$= (-4\overline{i} + 4\overline{j} + 4\overline{k}) – (4\overline{i} + 5\overline{j} + \overline{k})$$

$$= -8\overline{i} – \overline{j} + 3\overline{k}$$

$$\text{Now, } [\overline{PQ} \ \overline{PR} \ \overline{PS}] = \left| \begin{matrix} -4 & -6 & -2 \ -1 & 4 & 3 \ -8 & -1 & 3 \end{matrix} \right|$$

$$= [-4(12 + 3) + 6(-3 + 24) – 2(1 + 32)]$$

$$= -4(15) + 6(21) – 2(33)$$

$$= -60 + 126 – 66 = 0$$

$$\text{So, } \overline{PQ}, \overline{PR}, \overline{PS} \text{ are coplanar}$$

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SAQ-4 : If the points whose position vectors are 3¯i – 2¯j – ¯k, 2¯i + 3¯j – 4¯k, -¯i + ¯j + 2¯k, 4¯i + 5¯j + λ¯k are coplanar, then show that λ = -146/17

$$\overline{OP} = 3\overline{i} – 2\overline{j} – \overline{k}$$

$$\overline{OQ} = 2\overline{i} + 3\overline{j} – 4\overline{k}$$

$$\overline{OR} = -\overline{i} + \overline{j} + 2\overline{k}$$

$$\overline{OS} = 4\overline{i} + 5\overline{j} + \lambda\overline{k}$$

$$\text{where } O \text{ is the origin}$$

$$\overline{PQ} = \overline{OQ} – \overline{OP}$$

$$= (2\overline{i} + 3\overline{j} – 4\overline{k}) – (3\overline{i} – 2\overline{j} – \overline{k})$$

$$= -\overline{i} + 5\overline{j} – 3\overline{k}$$

$$\overline{PR} = \overline{OR} – \overline{OP}$$

$$= (-\overline{i} + \overline{j} + 2\overline{k}) – (3\overline{i} – 2\overline{j} – \overline{k})$$

$$= -4\overline{i} + 3\overline{j} + 3\overline{k}$$

$$\overline{PS} = \overline{OS} – \overline{OP}$$

$$= (4\overline{i} + 5\overline{j} + \lambda\overline{k}) – (3\overline{i} – 2\overline{j} – \overline{k})$$

$$= \overline{i} + 7\overline{j} + (\lambda + 1)\overline{k}$$

$$\text{But } [\overline{PQ} \ \overline{PR} \ \overline{PS}] = 0$$

$$= \left| \begin{matrix} -1 & 5 & -3 \ -4 & 3 & 3 \ 1 & 7 & \lambda + 1 \end{matrix} \right| = 0$$

$$= (-1)[3(\lambda + 1) – 21] – 5[-4(\lambda + 1) – 3] – 3[-28 – 3] = 0$$

$$= -1(3\lambda – 18) – 5(-4\lambda – 7) – 3(-31) = 0$$

$$= -3\lambda + 18 + 20\lambda + 35 + 93 = 0$$

$$= -3\lambda + 20\lambda + 35 + 93 + 18 = 0$$

$$= 17\lambda + 146 = 0$$

$$= 17\lambda = -146$$

$$\lambda = \frac{-146}{17}$$

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SAQ-5 : If ¯a , ¯b, ¯c are non-coplanar then prove that the points with position vectors 2¯a + 3¯b – ¯c, ¯a – 2¯b + 3¯c, 3¯a + 4¯b – 2¯c, ¯a – 6¯b + 6¯c are coplanar

$$\overline{OP} = 2\overline{a} + 3\overline{b} – \overline{c}$$

$$\overline{OQ} = \overline{a} – 2\overline{b} + 3\overline{c}$$

$$\overline{OR} = 3\overline{a} + 4\overline{b} – 2\overline{c}$$

$$\overline{OS} = \overline{a} – 6\overline{b} + 6\overline{c}$$

$$\text{where } O \text{ is the origin}$$

$$\overline{PQ} = \overline{OQ} – \overline{OP}$$

$$= (\overline{a} – 2\overline{b} + 3\overline{c}) – (2\overline{a} + 3\overline{b} – \overline{c})$$

$$= -\overline{a} – 5\overline{b} + 4\overline{c}$$

$$\overline{PR} = \overline{OR} – \overline{OP}$$

$$= (3\overline{a} + 4\overline{b} – 2\overline{c}) – (2\overline{a} + 3\overline{b} – \overline{c})$$

$$= \overline{a} + \overline{b} – \overline{c}$$

$$\overline{PS} = \overline{OS} – \overline{OP}$$

$$= (\overline{a} – 6\overline{b} + 6\overline{c}) – (2\overline{a} + 3\overline{b} – \overline{c})$$

$$= -\overline{a} – 9\overline{b} + 7\overline{c}$$

$$\text{Now, } [\overline{PQ} \ \overline{PR} \ \overline{PS}] = \left| \begin{matrix} -1 & -5 & 4 \ 1 & 1 & -1 \ -1 & -9 & 7 \end{matrix} \right|$$

$$= [-1(7 – 9) + 5(7 – 1) + 4(-9 + 1)][\overline{a} \ \overline{b} \ \overline{c}]$$

$$= [-1(-2) + 5(6) + 4(-8)][\overline{a} \ \overline{b} \ \overline{c}]$$

$$= [2 + 30 – 32][\overline{a} \ \overline{b} \ \overline{c}]$$

$$= 0 \times [\overline{a} \ \overline{b} \ \overline{c}] = 0$$

$$\text{So, } \overline{PQ}, \overline{PR}, \overline{PS} \text{ are coplanar}$$

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SAQ-6 : IF ABCDEF is a regular hexagon with centre O, then prove that ¯AB + ¯AC + ¯AD + ¯AE + ¯AF = 3¯AD = 6¯AO

$$\text{Given ABCDEF is a regular hexagon with centre O}$$

$$(\overline{AB} + \overline{AC}) + (\overline{AD}) + (\overline{AE} + \overline{AF})$$

$$= (\overline{AB} + \overline{AC}) + (\overline{AD}) + (\overline{BD} + \overline{CD})$$

$$= (\overline{AB} + \overline{BD}) + \overline{AD} + (\overline{AC} + \overline{CD})$$

$$= (\overline{AD}) + (\overline{AD}) + \overline{AD} = 3\overline{AD}$$

$$= 3(2\overline{AO})$$

$$= 6\overline{AO}$$

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SAQ-7 : Show that the points A(2¯i – ¯j + ¯k), B(¯i – 3¯j – 5¯k), C(3¯i – 4¯j – 4¯k) are the vertices of a right angled triangle

$$\text{We take}$$

$$\overline{OA} = (2\overline{i} – \overline{j} + \overline{k})$$

$$\overline{OB} = (\overline{i} – 3\overline{j} – 5\overline{k})$$

$$\overline{OC} = (3\overline{i} – 4\overline{j} – 4\overline{k})$$

$$\text{Where } O \text{ is the origin}$$

$$\overline{AB} = \overline{OB} – \overline{OA}$$

$$= (\overline{i} – 3\overline{j} – 5\overline{k}) – (2\overline{i} – \overline{j} + \overline{k})$$

$$= -\overline{i} – 2\overline{j} – 6\overline{k}$$

$$|\overline{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2}$$

$$= \sqrt{1 + 4 + 36}$$

$$= \sqrt{41}$$

$$\overline{BC} = \overline{OC} – \overline{OB}$$

$$= (3\overline{i} – 4\overline{j} – 4\overline{k}) – (\overline{i} – 3\overline{j} – 5\overline{k})$$

$$= 2\overline{i} – \overline{j} + \overline{k}$$

$$|\overline{BC}| = \sqrt{2^2 + (-1)^2 + 1^2}$$

$$= \sqrt{4 + 1 + 1}$$

$$= \sqrt{6}$$

$$\overline{CA} = \overline{OA} – \overline{OC}$$

$$= (2\overline{i} – \overline{j} + \overline{k}) – (3\overline{i} – 4\overline{j} – 4\overline{k})$$

$$= -\overline{i} + 3\overline{j} + 5\overline{k}$$

$$|\overline{CA}| = \sqrt{(-1)^2 + 3^2 + 5^2}$$

$$= \sqrt{1 + 9 + 25}$$

$$= \sqrt{35}$$

$$\text{Here } |\overline{AB}|^2 = 41; \quad |\overline{BC}|^2 + |\overline{CA}|^2$$

$$= 6 + 35$$

$$= 41$$

$$|\overline{AB}|^2 = |\overline{BC}|^2 + |\overline{CA}|^2$$

$$\text{A, B, C are the vertices of a right angled triangle}$$

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SAQ-8 : Show that the points (5, -1, 1), (7, -4, 7), (1, -6, 10) and (-1, -3, 4) are the vertices of a rhombus

$$\text{We take}$$

$$\overline{OA} = 5\overline{i} – \overline{j} + \overline{k}$$

$$\overline{OB} = 7\overline{i} – 4\overline{j} + 7\overline{k}$$

$$\overline{OC} = \overline{i} – 6\overline{j} + 10\overline{k}$$

$$\overline{OD} = -\overline{i} – 3\overline{j} + 4\overline{k}$$

$$\text{Where } O \text{ is the origin}$$

$$\overline{AB} = \overline{OB} – \overline{OA}$$

$$= (7\overline{i} – 4\overline{j} + 7\overline{k}) – (5\overline{i} – \overline{j} + \overline{k})$$

$$= 2\overline{i} – 3\overline{j} + 6\overline{k}$$

$$|\overline{AB}| = \sqrt{2^2 + (-3)^2 + 6^2}$$

$$= \sqrt{4 + 9 + 36}$$

$$= \sqrt{49}$$

$$= 7$$

$$\overline{BC} = \overline{OC} – \overline{OB}$$

$$= (\overline{i} – 6\overline{j} + 10\overline{k}) – (7\overline{i} – 4\overline{j} + 7\overline{k})$$

$$= -6\overline{i} – 2\overline{j} + 3\overline{k}$$

$$|\overline{BC}| = \sqrt{(-6)^2 + (-2)^2 + 3^2}$$

$$= \sqrt{36 + 4 + 9}$$

$$= \sqrt{49}$$

$$= 7$$

$$\overline{CD} = \overline{OD} – \overline{OC}$$

$$= (-\overline{i} – 3\overline{j} + 4\overline{k}) – (\overline{i} – 6\overline{j} + 10\overline{k})$$

$$= -2\overline{i} + 3\overline{j} – 6\overline{k}$$

$$|\overline{CD}| = \sqrt{(-2)^2 + 3^2 + (-6)^2}$$

$$= \sqrt{4 + 9 + 36}$$

$$= \sqrt{49}$$

$$= 7$$

$$\overline{DA} = \overline{OA} – \overline{OD}$$

$$= (5\overline{i} – \overline{j} + \overline{k}) – (-\overline{i} – 3\overline{j} + 4\overline{k})$$

$$= 6\overline{i} + 2\overline{j} – 3\overline{k}$$

$$|\overline{DA}| = \sqrt{6^2 + 2^2 + (-3)^2}$$

$$= \sqrt{36 + 4 + 9}$$

$$= \sqrt{49}$$

$$= 7$$

$$|\overline{AB}| = |\overline{BC}| = |\overline{CD}| = |\overline{DA}| \text{ so all 4 sides are equal}$$

$$\text{Also, }
\overline{AC} = \overline{OC} – \overline{OA}$$

$$= (\overline{i} – 6\overline{j} + 10\overline{k}) – (5\overline{i} – \overline{j} + \overline{k})$$

$$= -4\overline{i} – 5\overline{j} + 9\overline{k}$$

$$|\overline{AC}| = \sqrt{(-4)^2 + (-5)^2 + 9^2}$$

$$= \sqrt{16 + 25 + 81}$$

$$= \sqrt{122}$$

$$\overline{BD} = \overline{OD} – \overline{OB}$$

$$= (-\overline{i} – 3\overline{j} + 4\overline{k}) – (7\overline{i} – 4\overline{j} + 7\overline{k})$$

$$= -8\overline{i} + \overline{j} – 3\overline{k}$$

$$|\overline{BD}| = \sqrt{(-8)^2 + 1^2 + (-3)^2}$$

$$= \sqrt{64 + 1 + 9}$$

$$= \sqrt{74}$$

$$\text{Here } |\overline{AC}| \neq |\overline{BD}|$$

$$\text{So, A, B, C, D form a rhombus}$$

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SAQ-9 : If ¯a, ¯b, ¯c are noncoplanar, find the point of intersection of the line passing through the points 2¯a + 3¯b – ¯c, 3¯a + 4¯b -2¯c with the line joining the points ¯a – 2¯b + 3¯c, ¯a – 6¯b + 6¯c

$$\text{Let } P = 2\overline{a} + 3\overline{b} – \overline{c}$$

$$Q = 3\overline{a} + 4\overline{b} – 2\overline{c}$$

$$\text{Vector equation of the line passing through the points } P \text{ and } Q \text{ is}$$

$$\overline{r} = (1 – t)(2\overline{a} + 3\overline{b} – \overline{c}) + t(3\overline{a} + 4\overline{b} – 2\overline{c})$$

$$t \in \mathbb{R}$$

$$\overline{r} = (2 + t)\overline{a} + (3 + t)\overline{b} + (-1 – t)\overline{c} \quad \text{…(I)}$$

$$\text{Let } T = (\overline{a} – 2\overline{b} + 3\overline{c}), \quad S = (\overline{a} – 6\overline{b} + 6\overline{c})$$

$$\text{Vector equation of the line passing through } T \text{ and } S \text{ is}$$

$$\overline{r} = (1 – s)(\overline{a} – 2\overline{b} + 3\overline{c}) + s(\overline{a} – 6\overline{b} + 6\overline{c})$$

$$s \in \mathbb{R}$$

$$\overline{r} = \overline{a} + (-2 – 4s)\overline{b} + (3 + 3s)\overline{c} \quad \text{…(II)}$$

$$\text{If the two lines intersect at } P(\overline{r}) \text{ then from (I) and (II) we have}$$

$$(2 + t)\overline{a} + (3 + t)\overline{b} + (-1 – t)\overline{c} = \overline{a} + (-2 – 4s)\overline{b} + (3 + 3s)\overline{c}$$

$$\text{We compare the coefficients of } \overline{a}, \overline{b}, \overline{c} \text{ as } \overline{a}, \overline{b}, \overline{c} \text{ are non-coplanar}$$

$$2 + t = 1 \quad \text{…(1)}$$

$$3 + t = -2 – 4s \quad \text{…(2)}$$

$$-1 – t = 3 + 3s \quad \text{…(3)}$$

$$\text{From (1), } 2 + t = 1$$

$$t = -1$$

$$\text{From (2), } 3 + t = -2 – 4s$$

$$3 – 1 = -2 – 4s$$

$$4s = -4$$

$$s = -1$$

$$\text{Now substituting } t = -1 \text{ and } s = -1 \text{ in (I) we get point of intersection of the lines as } \overline{a} + 2\overline{b}$$

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SAQ-10 : Show that the line joining the pair of points 6¯a – 4¯b + 4¯c, -4¯c and the line joining the pair of points -¯a – 2¯b – 3¯c, ¯a + 2¯b – 5¯c intersect at the point -4¯c, when ¯a, ¯b, ¯c are non-coplanar vectors

$$\text{Let } P = -4\overline{c}$$

$$Q = 6\overline{a} – 4\overline{b} + 4\overline{c}$$

$$\text{Vector equation of the line joining the points } P \text{ and } Q \text{ is}$$

$$\overline{r} = (1 – t)(-4\overline{c}) + t(6\overline{a} – 4\overline{b} + 4\overline{c})$$

$$t \in \mathbb{R}$$

$$\overline{r} = (6t)\overline{a} – (4t)\overline{b} + (8t – 4)\overline{c} \quad \text{…(1)}$$

$$\text{Let } T = -\overline{a} – 2\overline{b} – 3\overline{c} \text{ and } S = \overline{a} + 2\overline{b} – 5\overline{c}$$

$$\text{Vector equation of the line joining the points } T \text{ and } S \text{ is}$$

$$\overline{r} = (1 – s)(-\overline{a} – 2\overline{b} – 3\overline{c}) + s(\overline{a} + 2\overline{b} – 5\overline{c})$$

$$s \in \mathbb{R}$$

$$\overline{r} = (2s – 1)\overline{a} + (4s – 2)\overline{b} + (-2s – 3)\overline{c} \quad \text{…(2)}$$

$$\text{If the two lines intersect each other at } P(\overline{r}) \text{ then from (1) and (2) we have}$$

$$(6t)\overline{a} – (4t)\overline{b} + (8t – 4)\overline{c} = (2s – 1)\overline{a} + (4s – 2)\overline{b} + (-2s – 3)\overline{c}$$

$$\text{Equating the corresponding coefficients of } \overline{a}, \overline{b}, \overline{c} \text{ we have}$$

$$6t = 2s – 1$$

$$6t – 2s = -1 \quad \text{…(3)}$$

$$-4t = 4s – 2$$

$$4t + 4s = 2 \quad \text{…(4)}$$

$$8t – 4 = -2s – 3$$

$$8t + 2s = 1 \quad \text{…(5)}$$

$$\text{Adding (3) and (5) we get}$$

$$14t + 0 = 0$$

$$14t = 0$$

$$t = 0$$