Scottish Highers Maths Vectors: Vectors **u** = $si + 2j

Vectors **u** = $si + 2j - k$ and **v** = $-3i + ti - 6k$ are perpendicular - Scottish Highers Maths - Question 1 - 2015

Question 1

Vectors **u** = $si + 2j - k$ and **v** = $-3i + ti - 6k$ are perpendicular. Determine the value of $t$.

Worked Solution & Example Answer:Vectors u = $si + 2j - k$ and v = $-3i + ti - 6k$ are perpendicular - Scottish Highers Maths - Question 1 - 2015

Step 1

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Answer

To determine the value of $t$ that makes vectors u and v perpendicular, we start with the scalar product (dot product) of the two vectors.

The scalar product of vectors u and v is given by:

$extbf{u} ullet extbf{v} = (si + 2j - k) ullet (-3i + ti - 6k)$

Calculating this step-by-step:

Multiply the corresponding components:
- For the i-component: $s imes -3 = -3s$
- For the j-component: $2 imes t = 2t$
- For the k-component: $-1 imes -6 = 6$
Combine these results:
$extbf{u} ullet extbf{v} = -3s + 2t + 6$
Since the vectors are perpendicular, their scalar product equals zero: $-3s + 2t + 6 = 0$

Step 2

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Answer

Now, we isolate $t$ in the equation:

$2t = 3s - 6$ $t = rac{3s - 6}{2}$

To find the specific value of $t$ , we need the value of $s$ . Assuming $s = 2$ (as derived from the marking scheme), we substitute:

$t = rac{3(2) - 6}{2} = rac{6 - 6}{2} = 0$

Thus the value of $t$ is: $t = 0$ .