Hướng dẫn giải Bài 6 (Trang 92 SGK Toán Hình học 12)

6. Cho mặt phẳng <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>α</mi></mfenced></math> có phương trình <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mi>x</mi><mo>+</mo><mn>5</mn><mi>y</mi><mo>-</mo><mi>z</mi><mo>-</mo><mn>2</mn><mo>=</mo><mn>0</mn></math> và đường thẳng <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math> có phương trình: <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="{" close=""><mtable><mtr><mtd><mi>x</mi><mo>=</mo><mn>12</mn><mo>+</mo><mn>4</mn><mi>t</mi></mtd></mtr><mtr><mtd><mi>y</mi><mo>=</mo><mn>9</mn><mo>+</mo><mn>3</mn><mi>t</mi></mtd></mtr><mtr><mtd><mi>z</mi><mo>=</mo><mn>1</mn><mo>+</mo><mi>t</mi></mtd></mtr></mtable></mfenced></math> a) Tìm giao điểm <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>M</mi></math> của đường thẳng <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math> và mặt phẳng <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>α</mi></mfenced></math>. b) Viết phương trình mặt phẳng <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>β</mi></mfenced></math> chứa điểm <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>M</mi></math> và vuông góc với đường thẳng <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math> Giải: a) Thay <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mn>12</mn><mo>+</mo><mn>4</mn><mi>t</mi><mo>,</mo><mo> </mo><mi>y</mi><mo>=</mo><mn>9</mn><mo>+</mo><mn>3</mn><mi>t</mi><mo>,</mo><mo> </mo><mi>z</mi><mo>=</mo><mn>1</mn><mo>+</mo><mi>t</mi></math> vào phương trình mp<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>α</mi></mfenced></math> ta được <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mfenced><mrow><mn>12</mn><mo>+</mo><mn>4</mn><mi>t</mi></mrow></mfenced><mo>+</mo><mn>5</mn><mfenced><mrow><mn>9</mn><mo>+</mo><mn>3</mn><mi>t</mi></mrow></mfenced><mo>-</mo><mfenced><mrow><mn>1</mn><mo>+</mo><mi>t</mi></mrow></mfenced><mo>-</mo><mn>2</mn><mo>=</mo><mn>0</mn><mo>⇔</mo><mn>26</mn><mi>t</mi><mo>+</mo><mn>78</mn><mo>=</mo><mn>0</mn><mo>⇔</mo><mi>t</mi><mo>=</mo><mo>-</mo><mn>3</mn></math> Khi đó <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mi>y</mi><mo>=</mo><mn>0</mn><mo>,</mo><mo> </mo><mi>z</mi><mo>=</mo><mo>-</mo><mn>2</mn><mo>.</mo></math> Vậy <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math> cắt <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>α</mi></mfenced></math> tại điểm <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>M</mi><mfenced><mrow><mn>0</mn><mo>;</mo><mo> </mo><mn>0</mn><mo>;</mo><mo> </mo><mo>-</mo><mn>2</mn></mrow></mfenced></math>. b) Đường thẳng <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math> có vecto chỉ phương là <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><msub><mi>a</mi><mi>d</mi></msub><mo>→</mo></mover><mo>=</mo><mfenced><mrow><mn>4</mn><mo>;</mo><mo> </mo><mn>3</mn><mo>;</mo><mo> </mo><mn>1</mn></mrow></mfenced></math> Mp<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>β</mi></mfenced></math> vuông góc với <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math> thì <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>α</mi></mfenced></math> có vecto pháp tuyến <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>n</mi><mo>→</mo></mover><mo>=</mo><mover><msub><mi>a</mi><mi>d</mi></msub><mo>→</mo></mover><mo>=</mo><mfenced><mrow><mn>4</mn><mo>;</mo><mo> </mo><mn>3</mn><mo>;</mo><mo> </mo><mn>1</mn></mrow></mfenced></math> nên có phương trình là <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mfenced><mrow><mi>x</mi><mo>-</mo><mn>0</mn></mrow></mfenced><mo>+</mo><mn>3</mn><mfenced><mrow><mi>y</mi><mo>-</mo><mn>0</mn></mrow></mfenced><mo>+</mo><mn>1</mn><mfenced><mrow><mi>z</mi><mo>+</mo><mn>2</mn></mrow></mfenced><mo>=</mo><mn>0</mn><mo>⇔</mo><mn>4</mn><mi>x</mi><mo>+</mo><mn>3</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>+</mo><mn>2</mn><mo>=</mo><mn>0</mn><mo>.</mo></math>