Question

if a=2i-3j+5k, b-i+4j+2k then find (a+b)×(a-b) and unit vector perpendicular to both a+b and a-b​

Answer 1

Step-by-step explanation:

a=2i–3j+k

b=i+4j–2k

(a+b)×(a–b)

(a+b)= 2i–3j+k+i+4j–2k

=3i+j–k×i–7j+3k

(a–b)=2i–3j+k–(i+4j–2k)

=i–7j+3k

(a+b)×(a–b)=(3i+j–k)×(i–7j+3k)

i j. k

3. 1. –1

1. –7. 3

=–4i–10j–22k

=–2(2i+5j+11k)

Answer 2

The value of (a+b)×(a-b) is -4i -10j -22k and unit vector perpendicular to both (a+b) and (a-b​) is $\frac{(2i+5j+11k)}{242}$

Given:

Two vectors: a = 2i-3j+5k and b = i+4j+2k

To Find:

The value of (a+b)×(a-b) and unit vector perpendicular to both (a+b) and

(a-b​).

Solution:

We have,

a = 2i-3j+5k

b = i+4j+2k

From the knowledge of Vector addition and Vector subtraction, first, we need to find the values of (a+b) and (a-b) separately.

(a+b) = (2i-3j+5k) + (i+4j+2k)

        = 3i+j–k

(a-b) = (2i-3j+5k) - (i+4j+2k)

        = i–7j+3k

Now following vector multiplication rule and let,

c = (a+b) × (a-b)

  = (3i+j–k) × (i–7j+3k)

  = $\left[\begin{array}{ccc}i&j&k\\3&1&-1\\1&-7&3\end{array}\right]$ [Following matrix form]

  = i [(1 × 3) - (-7 × -1)] - j [(3 × 3) - (1 × -1)] + k [(-7 × 3) - (1 × 1)]

  = i [3 - 7] - j [9 + 1] + k [-21 - 1]

  = -4i -10j -22k

We have the value of (a+b) × (a-b), which is again needed for finding the unit vector of (a+b) and (a-b)

Basically, we have to find the unit vector of c.

therefore,

The unit vector of c  

= $\frac{1}{Magnitude of c}$ ×c

= $\frac{1}{\sqrt{(-4)^{2} +(-10)^{2} +(-22)^{2} } }$ × (-4i -10j -22k)

= $\frac{1}{484}$ × (-4i -10j -22k)

= $\frac{1}{484}$ × –2(2i+5j+11k)

= $\frac{1}{242}$ × (2i+5j+11k)

= $\frac{(2i+5j+11k)}{242}$

Therefore, the value of (a+b)×(a-b) is -4i -10j -22k and unit vector perpendicular to both (a+b) and (a-b​) is $\frac{(2i+5j+11k)}{242}$

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