Mathematics Lab Activity-12 Class XII

By CBSE MATHEMATICS February 22, 2024

Mathematics Lab Activity-12 Class XII

Mathematics Laboratory Activities 12 on Vector Algebra for class XI Non-Medical students with complete observation tables strictly according to the CBSE syllabus.

Chapter - 10

vector algebra

Activity - 12

Objective

To verify geometrically that : $\vec{c}\times(\vec{a}\times\vec{b})=\vec{c}\times\vec{a}+\vec{c}\times\vec{b}$

Material Required

Geometry box, cardboard, white paper, cutter, sketch pen, cello tape etc.

Procedure

1. Fix a white paper on the cardboard.

2. Draw a line segment OA (=6 cm, say) and let it represent $\vec{c}$ .

3. Draw another line segment OB (= 4 cm, say) at an angle (say 60^o) with OA. Let $\overrightarrow{OB}=\vec{a}$ .

Figure 12.1

4. Draw BC (= 3cm) making an angle (say 30^o) with $\overrightarrow{OA}$ . Let $\overrightarrow{BC}=\vec{b}$

5. Draw perpendicular BM, CL and BN and complete the parallelograms OAPC, OAQB and BQPC.

6. $\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{BC}=\vec{a}+\vec{b}$ , and let $\angle COA=\alpha$

7. Area of parallelogram OAQB = $|\vec{c}\times\vec{a}|$

Area of parallelogram BQPC = $|\vec{c}\times\vec{b}|$

Area of parallelogram OAPC = $|\vec{c}\times(\vec{a}+\vec{b})|$

= (OA)(CL)

= (OA)(LN + NC)

= (OA)(BM + NC)

= (OA)(BM) + (OA)(NC)

= Area of parallelogram OAQB + Area of parallelogram BQPC

= $|\vec{c}\times\vec{a}|+|\vec{c}\times\vec{b}|$

Observations

In triangle BOM in figure 12.2
$\frac{BM}{OB}=sin\:60^{o}$

$\frac{BM}{4}=\frac{\sqrt{3}}{2}$

$\Rightarrow BM=2\sqrt{3}=2\times 1.73=3.46\:cm$

In figure 12.1 Area of Parallelogram OAQB = 6 ✕ 3.46 = 20.76 cm²

Figure 12.2

Figure 12.3

In triangle BCN in figure 12.3
$\frac{CN}{BC}=sin\:30^{o}$

$\frac{CN}{3}=\frac{1}{2}$

⇒ CN = 3/2 = 1.5 cm

In figure 12.1

Area of Parallelogram BQPC = 6 ✕ 1.5 cm = 9 cm²

Now Area of Parallelogram OAQB + Area of Parallelogram BQPC

= 20.76 cm² + 9 cm²

= 29.76 9 cm² ........ (i)

In Figure 12.1

Base of Parallelogram OAPC = 6 cm

Height CL = BM + CN = 3.46 + 1.5 = 4.96 cm

Area of parallelogram OAPC = 6 ✕ 4.96 = 29.76 cm² ...... (ii)

From (i) and (ii) we conclude that

Area of parallelogram OAPC = Area of Parallelogram OAQB + Area of Parallelogram BQPC

In vector form this can be written as

⇒ $|\vec{c}\times(\vec{a}\times\vec{b})|=|\vec{c}\times\vec{a}|+|\vec{c}\times\vec{b}|$

$\vec{c}\times(\vec{a}\times\vec{b})=\vec{c}\times\vec{a}+\vec{c}\times\vec{b}$

Result

Through this activity we prove that $\vec{c}\times(\vec{a}\times\vec{b})=\vec{c}\times\vec{a}+\vec{c}\times\vec{b}$

Applications

Through this activity, distributive property of vector multiplication over addition can be explained.

VIVA – VOICE

Q1. Is   $\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$ always ?
Ans. Yes.

Q2. Can we write $\vec{a}.(\vec{b}+\vec{c})=\vec{a}\:.\:\vec{b}+\vec{a}\:.\:\vec{c}$
Ans. Yes

Q3. What does $|\vec{a}\times\vec{b}|$ represents ?
Ans. It represents the area of parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ .

Q4. What does   $\frac{1}{2}|\vec{a}\times\vec{b}|$ represent ?
Ans. It represents area of triangle whose sides are $\vec{a}$ and $\vec{b}$ .

Q5. What is condition for collinear vectors ?
Ans.   $\vec{a}\times\vec{b}=\vec{0}$

Q6. Define cross product of vectors $\vec{a}$ and $\vec{b}$ .
Ans. $\vec{a}\times\vec{b}=ab\:sin\theta\:\hat{n}$ , where 0 ≤ θ ≤ π and $\hat{n}$ is unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ .

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