Given vector a=3i+3j-2k,b=-i+3j+4k and c=4i-2j-6k. Find a unit vector coplanar with a and b but perpendicular to c

To find a unit vector that is coplanar with vectors **a** and **b** but perpendicular to vector **c**, we can follow a systematic approach. First, we need to determine a vector that lies in the plane formed by **a** and **b**. This can be achieved by taking the cross product of **a** and **b**. Then, we will find a vector that is perpendicular to **c** and lies in the same plane. Finally, we will normalize this vector to obtain the unit vector.

Step 1: Define the Vectors

We have the following vectors:

• a = 3i + 3j - 2k
• b = -i + 3j + 4k
• c = 4i - 2j - 6k

Step 2: Calculate the Cross Product of a and b

The cross product of two vectors results in a vector that is perpendicular to both. We can calculate the cross product **a × b** using the determinant of a matrix:

| i   j   k   |
| 3   3  -2   |
| -1  3   4   |

Calculating this determinant, we have:

= i(3*4 - (-2)*3) - j(3*4 - (-2)*(-1)) + k(3*3 - 3*(-1))
= i(12 + 6) - j(12 - 2) + k(9 + 3)
= 18i - 10j + 12k

Thus, the cross product **a × b** = 18i - 10j + 12k.

Step 3: Find a Vector Perpendicular to c

Next, we need to find a vector that is perpendicular to **c**. A vector **v** is perpendicular to **c** if their dot product is zero:

Let **v** = 18i - 10j + 12k. We need to ensure:

v • c = 0

Calculating the dot product:

(18)(4) + (-10)(-2) + (12)(-6) = 72 + 20 - 72 = 20

Since the dot product is not zero, **v** is not perpendicular to **c**. We need to find a linear combination of **a** and **b** that is perpendicular to **c**. Let’s express a general vector in the plane formed by **a** and **b**:

v = k1 * a + k2 * b

Step 4: Set Up the Perpendicular Condition

We can express **v** as:

v = k1(3i + 3j - 2k) + k2(-i + 3j + 4k)

Now, we can expand this:

v = (3k1 - k2)i + (3k1 + 3k2)j + (-2k1 + 4k2)k

Now we need to ensure that this vector is perpendicular to **c**:

(3k1 - k2)(4) + (3k1 + 3k2)(-2) + (-2k1 + 4k2)(-6) = 0

Step 5: Solve the Equation

Expanding this gives:

12k1 - 4k2 - 6k1 - 6k2 + 12k1 - 24k2 = 0

Combining like terms:

(12k1 - 6k1 + 12k1) + (-4k2 - 6k2 - 24k2) = 0

18k1 - 34k2 = 0

From this, we can express k1 in terms of k2:

k1 = (34/18)k2 = (17/9)k2

Step 6: Choose k2 and Find the Vector

Let’s choose k2 = 9 (to simplify calculations), then:

k1 = 17

Now substituting back into the expression for **v**:

v = 17(3i + 3j - 2k) + 9(-i + 3j + 4k)
= (51 - 9)i + (51 + 27)j + (-34 + 36)k
= 42i + 78j + 2k

Step 7: Normalize the Vector

To find the unit vector, we need to normalize **v**:

|v| = √(42^2 + 78^2 + 2^2) = √(1764 + 6084 + 4) = √7848

Calculating this gives:

|v| = 88.5

Thus, the unit vector **u** is:

u = (1/|v|) * v = (1/88.5)(42i + 78j + 2k)

Finally, we can express the unit vector:

u ≈ 0.474i + 0.883j + 0.023k

Final Result

The unit vector that is coplanar with **a** and **b** but perpendicular to **c** is approximately:

u ≈ 0.474i + 0.883j + 0.023k

This process illustrates how to manipulate vectors in three-dimensional space to find specific relationships, such as coplanarity and perpendicularity.