Question text 2Marks Let [math: p=(x+2)i+xj+k]\textbf{p}=(x+2)\textbf{i}+x\textbf{j}+\textbf{k} and [math: q=−2i+xj+5k]\textbf{q}=-2\textbf{i}+x\textbf{j}+5\textbf{k}. Find [math: x] such that [math: p]\textbf{p} is perpendicular to [math: q]\textbf{q}.Thus [math: x=] Answer 1[input].Notes Report question issue Question 14 Notes多项填空题
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Question

Question text 2Marks Let [math: p=(x+2)i+xj+k]\textbf{p}=(x+2)\textbf{i}+x\textbf{j}+\textbf{k} and [math: q=−2i+xj+5k]\textbf{q}=-2\textbf{i}+x\textbf{j}+5\textbf{k}. Find [math: x] such that [math: p]\textbf{p} is perpendicular to [math: q]\textbf{q}.Thus [math: x=] Answer 1[input].Notes Report question issue Question 14 Notes多项填空题

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BlackTom AI · Accepted Answer

To determine the value of x, we interpret the condition that p and q are perpendicular as their dot product being zero: p · q = 0.

First, write the vectors: p = (x+2, x, 1) and q = (-2, x, 5).
Compute the dot product: p · q = (x+2)(-2) + (x)(x) + (1)(5).
Simplify term by term: (x+2)(-2) = -2x - 4, x·x = x^2, and 1·5 = 5. So p · q = (-2x - 4) + x^2 + 5.
Combine like terms: p · q = x^2 - 2x + 1.
Recognize the quadratic: x^2 - 2x + 1 = (x - 1)^2.
Set the dot product to zero for perpendicularity: (x - 1)^2 = 0.
Solve for x: x = 1.
Thus the required value of x is 1. Note that if the problem intended p·p to be perpendicular to q·q, that interpretation would be incorrect in standard vector algebra, since those are scalar magnitudes and not vectors; however, the conventional and consistent interpretation here yields x = 1.